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# 10.2: The Hyperbola

The Hyperbola
In this section, you will:
• Locate a hyperbola’s vertices and foci.
• Write equations of hyperbolas in standard form.
• Graph hyperbolas centered at the origin.
• Graph hyperbolas not centered at the origin.
• Solve applied problems involving hyperbolas.

What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See [link].

<figure class="small" id="Figure_10_02_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"> <figcaption>A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom.</figcaption> </figure>

Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom.

# Locating the Vertices and Foci of a Hyperbola

In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See [link].

<figure class="small" id="Figure_10_02_002"> <figcaption>A hyperbola</figcaption> </figure>

Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) in a plane such that the difference of the distances between<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. Theconjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See [link].

<figure class="medium" id="Figure_10_02_003"> <figcaption>Key features of the hyperbola</figcaption> </figure>

In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

## Deriving the Equation of an Ellipse Centered at the Origin

Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −c,0 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] c,0 ) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) such that the difference of the distances from<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) to the foci is constant. See [link].

<figure class="small" id="Figure_10_02_013"></figure>

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a,0 ) is a vertex of the hyperbola, the distance from<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −c,0 ) to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a,0 ) is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>−</mo><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −c )=a+c. The distance from<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] c,0 ) to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a,0 ) is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>−</mo><mi>a</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The sum of the distances from the foci to the vertex is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] a+c )−( c−a )=2a

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) is a point on the hyperbola, we can define the following variables:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><msub><mi>d</mi></msub></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 2 =the distance from ( −c,0 ) to ( x,y ) d 1 =the distance from ( c,0 ) to ( x,y )

By definition of a hyperbola,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] d 2 − d 1  is constant for any point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) on the hyperbola. We know that the difference of these distances is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the vertex<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>a</mi><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]It follows that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] d 2 − d 1 =2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>                                      </mtext><msub/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] d 2 − d 1 = (x−(−c)) 2 + (y−0) 2 − (x−c) 2 + (y−0) 2 =2a Distance Formula (x+c) 2 + y 2 − (x−c) 2 +y 2 =2a Simplify expressions.                             (x+c) 2 + y 2 =2a+ (x−c) 2 + y 2 Move radical to opposite side.                              (x+c) 2 + y 2 = ( 2a+ (x−c) 2 + y 2 ) 2 Square both sides.                      x 2 +2cx+ c 2 + y 2 =4 a 2 +4a (x−c) 2 + y 2 + (x−c) 2 + y 2 Expand the squares.                      x 2 +2cx+ c 2 + y 2 =4 a 2 +4a (x−c) 2 + y 2 + x 2 −2cx+ c 2 +y 2 Expand remaining square.                                              2cx=4 a 2 +4a (x−c) 2 + y 2 −2cx Combine like terms.                                  4cx−4 a 2 =4a (x−c) 2 + y 2 Isolate the radical.                                       cx− a 2 =a (x−c) 2 + y 2Divide by 4.                                    ( cx− a 2 ) 2 = a 2 [ (x−c) 2 + y 2 ] 2 Square both sides.                      c 2 x 2 −2 a 2 cx+ a4 = a 2 ( x 2 −2cx+ c 2 + y 2 ) Expand the squares.                     c 2 x 2 −2 a 2 cx+ a 4 = a 2 x 2 −2 a 2 cx+ a 2 c 2 + a 2 y 2Distribute  a 2 .                                    a 4 + c 2 x 2 = a 2 x 2 + a 2 c 2 + a 2 y 2 Combine like terms.                   c 2 x 2 − a 2x 2 − a 2 y 2 = a 2 c 2 − a 4 Rearrange terms.                     x 2 ( c 2 − a 2 )− a 2 y 2 = a 2 ( c 2 − a 2 ) Factor common terms.                              x 2 b 2 − a 2 y 2 = a 2 b 2 Set  b 2 = c 2 − a 2 .                              x 2 b 2 a 2 b 2 − a 2 y 2 a 2 b 2 = a 2 b 2a 2 b 2 Divide both sides by  a 2 b 2                                      x 2 a 2 − y 2 b 2 =1

This equation defines a hyperbola centered at the origin with vertices<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±a,0 ) and co-vertices<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0±b ).

Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 ) and transverse axis on the x-axis is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 a 2 − y 2 b 2 =1

where

• the length of the transverse axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±a,0 )
• the length of the conjugate axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>b</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±b )
• the distance between the foci is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>c</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±c,0 )
• the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] b a x

The standard form of the equation of a hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 ) and transverse axis on the y-axis is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 a 2 − x 2 b 2 =1

where

• the length of the transverse axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±a )
• the length of the conjugate axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>b</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±b,0 )
• the distance between the foci is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>c</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±c )
• the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b x

Note that the vertices, co-vertices, and foci are related by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2 . When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

<figure id="Figure_10_02_004"> <figcaption>(a) Horizontal hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 ) (b) Vertical hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 )</figcaption> </figure>

Given the equation of a hyperbola in standard form, locate its vertices and foci.

1. Determine whether the transverse axis lies on the x- or y-axis. Notice that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2  is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
1. If the equation has the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1, then the transverse axis lies on the x-axis. The vertices are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mo>±</mo><mi>a</mi><mo>,</mo><mn>0</mn><mo stretchy="false">)</mo><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and the foci are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±c,0 ).
2. If the equation has the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 a 2 − x 2 b 2 =1, then the transverse axis lies on the y-axis. The vertices are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mo>±</mo><mi>a</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and the foci are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±c ).
2. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 .
3. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 .
Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 49 − x 2 32 =1.

The equation has the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 a 2 − x 2 b 2 =1, so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>0</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mn>1</mn><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] y 2 49 − x 2 32    1= y 2 49 − 0 2 32    1= y 2 49 y 2 =49    y=± 49 =±7

The foci are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±c ). Solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>c</mi><mo>=</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 = 49+32 = 81 =9

Therefore, the vertices are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±7 ), and the foci are located at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,9 ).

Identify the vertices and foci of the hyperbola with equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 9 − y 2 25 =1.

Vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±3,0 ); Foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ± 34 ,0 )

# Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

## Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 ),we see that the vertices, co-vertices, and foci are related by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2 . Note that this equation can also be rewritten as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 = c 2 − a 2 . This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

Given the vertices and foci of a hyperbola centered at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0, 0 ), write its equation in standard form.

1. Determine whether the transverse axis lies on the x- or y-axis.
1. If the given coordinates of the vertices and foci have the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±a,0 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±c,0 ), respectively, then the transverse axis is the x-axis. Use the standard form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1.
2. If the given coordinates of the vertices and foci have the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±a ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±c ), respectively, then the transverse axis is the y-axis. Use the standard form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 a 2 − x 2 b 2 =1.
2. Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2  using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 = c 2 − a 2 .
3. Substitute the values for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2  into the standard form of the equation determined in Step 1.
Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±6,0 ) and foci<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±2 10 ,0 )?

The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1.

The vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±6,0 ), so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mn>6</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2 =36.

The foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±2 10 ,0 ), so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mn>2</mn><msqrt/></mrow></annotation-xml></semantics>[/itex] 10  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 =40.

Solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 , we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>b</mi></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 = c 2 − a 2 b 2 =40−36 Substitute for  c 2  and  a 2 . b 2 =4 Subtract.

Finally, we substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2 =36 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 =4 into the standard form of the equation,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1.  The equation of the hyperbola is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 36 − y 2 4 =1, as shown in [link].

<figure class="small" id="Figure_10_02_014"></figure>

What is the standard form equation of the hyperbola that has vertices<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±2 ) and foci<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±2 5 )?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 4 − x 2 16 =1

## Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]units horizontally and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]units vertically, the center of the hyperbola will be<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ). This translation results in the standard form of the equation we saw previously, with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]replaced by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x−h ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]replaced by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] y−k ).

Standard Forms of the Equation of a Hyperbola with Center (h, k)

The standard form of the equation of a hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ) and transverse axis parallel to the x-axis is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( x−h ) 2 a 2 − ( y−k ) 2 b 2 =1

where

• the length of the transverse axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±a,k )
• the length of the conjugate axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>b</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±b )
• the distance between the foci is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>c</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±c,k )

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and its width is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>b</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]The slopes of the diagonals are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] b a ,and each diagonal passes through the center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ). Using the point-slope formula, it is simple to show that the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] b a ( x−h )+k. See [link]a

The standard form of the equation of a hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ) and transverse axis parallel to the y-axis is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y−k ) 2 a 2 − ( x−h ) 2 b 2 =1

where

• the length of the transverse axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±a )
• the length of the conjugate axis is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>b</mi></mrow></annotation-xml></semantics>[/itex]
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±b,k )
• the distance between the foci is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>c</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±c )

Using the reasoning above, the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b ( x−h )+k. See [link]b.

<figure id="Figure_10_02_005"> <figcaption>(a) Horizontal hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ) (b) Vertical hyperbola with center<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k )</figcaption> </figure>

Like hyperbolas centered at the origin, hyperbolas centered at a point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ) have vertices, co-vertices, and foci that are related by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 = a 2 + b 2 . We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

Given the vertices and foci of a hyperbola centered at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ),write its equation in standard form.

1. Determine whether the transverse axis is parallel to the x- or y-axis.
1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( x−h ) 2 a 2 − ( y−k ) 2 b 2 =1.
2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( y−k ) 2 a 2 − ( x−h ) 2 b 2 =1.
2. Identify the center of the hyperbola,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ),using the midpoint formula and the given coordinates for the vertices.
3. Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2  by solving for the length of the transverse axis,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi></mrow></annotation-xml></semantics>[/itex], which is the distance between the given vertices.
4. Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2  using<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]found in Step 2 along with the given coordinates for the foci.
5. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2  using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 = c 2 − a 2 .
6. Substitute the values for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>, </mo><mi>k</mi><mo>, </mo><msup><mi>a</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2  into the standard form of the equation determined in Step 1.
Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mn>−2</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mn>6</mn><mo>,</mo><mn>−2</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]and foci at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mn>−2</mn><mo>,</mo><mn>−2</mn><mo stretchy="false">)</mo></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mn>8</mn><mo>,</mo><mn>−2</mn><mo stretchy="false">)</mo><mo>?</mo></mrow></annotation-xml></semantics>[/itex]

The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−h ) 2 a 2 − ( y−k ) 2 b 2 =1

First, we identify the center,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k ). The center is halfway between the vertices<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,−2 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 6,−2 ). Applying the midpoint formula, we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k )=( 0+6 2 , −2+( −2 ) 2 )=( 3,−2 )

Next, we find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2 . The length of the transverse axis,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]is bounded by the vertices. So, we can find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2  by finding the distance between the x-coordinates of the vertices.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>2</mn><mi>a</mi><mo>=</mo><mrow><mo>|</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 0−6 | 2a=6   a=3 a 2 =9

Now we need to find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] c 2 . The coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±c,k ). So<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h−c,k )=( −2,−2 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h+c,k )=( 8,−2 ). We can use the x-coordinate from either of these points to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Using the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 8,−2 ), and substituting<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>=</mo><mn>3</mn><mo>,</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>h</mi><mo>+</mo><mi>c</mi><mo>=</mo><mn>8</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 3+c=8       c=5      c 2 =25

Next, solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2  using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 = c 2 − a 2 :

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mi>b</mi></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 = c 2 − a 2     =25−9     =16

Finally, substitute the values found for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>h</mi><mo>,</mo><mi>k</mi><mo>,</mo><msup/></mrow></annotation-xml></semantics>[/itex] a 2 ,and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2  into the standard form of the equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] (x−3) 2 9 − (y+2) 2 16 =1

What is the standard form equation of the hyperbola that has vertices<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1,−2 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1,8 ) and foci<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1,−10 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1,16 )?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y−3 ) 2 25 + ( x−1 ) 2 144 =1

# Graphing Hyperbolas Centered at the Origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1 for horizontal hyperbolas and the standard form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 a 2 − x 2 b 2 =1 for vertical hyperbolas.

Given a standard form equation for a hyperbola centered at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 ), sketch the graph.

1. Determine which of the standard forms applies to the given equation.
2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
1. If the equation is in the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1, then
• the transverse axis is on the x-axis
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±a,0 )
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±b )
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±c,0 )
• the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] b a x
2. If the equation is in the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 a 2 − x 2 b 2 =1, then
• the transverse axis is on the y-axis
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±a )
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±b,0 )
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±c )
• the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b x
3. Solve for the coordinates of the foci using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mo>±</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 .
4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.
Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form

Graph the hyperbola given by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 64 − x 2 36 =1.  Identify and label the vertices, co-vertices, foci, and asymptotes.

The standard form that applies to the given equation is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] y 2 a 2 − x 2 b 2 =1. Thus, the transverse axis is on the y-axis

The coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±a )=( 0,± 64 )=( 0,±8 )

The coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±b,0 )=( ± 36 , 0 )=( ±6,0 )

The coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±c ), where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mo>±</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 . Solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>c</mi><mo>=</mo><mo>±</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 =± 64+36 =± 100 =±10

Therefore, the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±10 )

The equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b x=± 8 6 x=± 4 3 x

Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in [link].

<figure class="medium" id="Figure_10_02_006"></figure>

Graph the hyperbola given by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 144 − y 2 81 =1. Identify and label the vertices, co-vertices, foci, and asymptotes.

vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±12,0 ); co-vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,±9 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] ±15,0 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 3 4 x;

# Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k )other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( x−h ) 2 a 2 − ( y−k ) 2 b 2 =1 for horizontal hyperbolas, and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( y−k ) 2 a 2 − ( x−h ) 2 b 2 =1 for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

Given a general form for a hyperbola centered at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h, k ), sketch the graph.

1. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.
1. If the equation is in the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( x−h ) 2 a 2 − ( y−k ) 2 b 2 =1, then
• the transverse axis is parallel to the x-axis
• the center is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k )
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±a,k )
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±b )
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±c,k )
• the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] b a ( x−h )+k
2. If the equation is in the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( y−k ) 2 a 2 − ( x−h ) 2 b 2 =1, then
• the transverse axis is parallel to the y-axis
• the center is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k )
• the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±a )
• the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±b,k )
• the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±c )
• the equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] a b ( x−h )+k
3. Solve for the coordinates of the foci using the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mo>±</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 .
4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.
Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form

Graph the hyperbola given by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −4 y 2 −36x−40y−388=0. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 9 x 2 −36x )−( 4 y 2 +40y )=388

Factor the leading coefficient of each expression.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>9</mn><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 −4x )−4( y 2 +10y )=388

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>9</mn><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x 2 −4x+4 )−4( y 2 +10y+25 )=388+36−100

Rewrite as perfect squares.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] ( x−2 ) 2 −4 ( y+5 ) 2 =324

Divide both sides by the constant term to place the equation in standard form.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 ) 2 36 − ( y+5 ) 2 81 =1

The standard form that applies to the given equation is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( x−h ) 2 a 2 − ( y−k ) 2 b 2 =1, where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2 =36 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 =81,or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mn>6</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>=</mo><mn>9.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Thus, the transverse axis is parallel to the x-axis. It follows that:

the center of the ellipse is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k )=( 2,−5 )
the coordinates of the vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±a,k )=( 2±6,−5 ), or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −4,−5 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 8,−5 )
the coordinates of the co-vertices are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h,k±b )=( 2,−5±9 ), or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,−14 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,4 )
the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] h±c,k ), where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mo>±</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] a 2 + b 2 . Solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex]we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>c</mi><mo>=</mo><mo>±</mo><msqrt/></mrow></annotation-xml></semantics>[/itex] 36+81 =± 117 =±3 13

Therefore, the coordinates of the foci are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2−3 13 ,−5 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2+3 13 ,−5 ).

The equations of the asymptotes are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] b a ( x−h )+k=± 3 2 ( x−2 )−5.

Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in [link].

<figure class="small" id="Figure_10_02_008"></figure>

Graph the hyperbola given by the standard form of an equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] ( y+4 ) 2 100 − ( x−3 ) 2 64 =1. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

center:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,−4 ); vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,−14 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,6 ); co-vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −5,−4 ); and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 11,−4 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,−4−2 41 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,−4+2 41 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>±</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 5 4 ( x−3 )−4

# Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See [link]. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

<figure class="small" id="Figure_10_02_010"> <figcaption>Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)</figcaption> </figure>

The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In [link] we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown in [link]. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.

<figure class="small" id="Figure_10_02_011"> <figcaption>Project design for a natural draft cooling tower</figcaption> </figure>

Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 a 2 − y 2 b 2 =1,where the branches of the hyperbola form the sides of the cooling tower. We must find the values of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2to complete the model.

First, we find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2 . Recall that the length of the transverse axis of a hyperbola is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]This length is represented by the distance where the sides are closest, which is given as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>65.3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]meters. So,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>a</mi><mo>=</mo><mn>60.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Therefore,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mn>30</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] a 2 =900.

To solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] b 2 ,we need to substitute for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]in our equation using a known point. To do this, we can use the dimensions of the tower to find some point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] 2 a 2 − y 2 b 2 =1 Standard form of horizontal hyperbola.            b 2 = y 2 x 2 a 2 −1 Isolate  b 2              = (79.6) 2 (36) 2900 −1 Substitute for  a 2 ,x, and y              ≈14400.3636 Round to four decimal places

The sides of the tower can be modeled by the hyperbolic equation

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 900 − y 2 14400.3636  =1,or   x 2 30 2 − y 2 120.0015 2   =1

A design for a cooling tower project is shown in [link]. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

<figure class="small" id="Figure_10_02_012"></figure>

The sides of the tower can be modeled by the hyperbolic equation.<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 400 − y 2 3600 =1or  x 2 20 2 − y 2 60 2 =1.

Access these online resources for additional instruction and practice with hyperbolas.

# Key Equations

 Hyperbola, center at origin, transverse axis on x-axis x[/itex] 2 a 2 − y 2 b 2 =1 Hyperbola, center at origin, transverse axis on y-axis y[/itex] 2 a 2 − x 2 b 2 =1 Hyperbola, center at(h,k),[/itex]transverse axis parallel to x-axis ([/itex] x−h ) 2 a 2 − ( y−k ) 2 b 2 =1 Hyperbola, center at(h,k),[/itex]transverse axis parallel to y-axis ([/itex] y−k ) 2 a 2 − ( x−h ) 2 b 2 =1

# Key Concepts

• A hyperbola is the set of all points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) in a plane such that the difference of the distances between<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) and the foci is a positive constant.
• The standard form of a hyperbola can be used to locate its vertices and foci. See [link].
• When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See [link] and [link].
• When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See [link] and [link].
• Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. See [link].

# Section Exercises

## Verbal

Define a hyperbola in terms of its foci.

A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (foci) is a positive constant.

What can we conclude about a hyperbola if its asymptotes intersect at the origin?

What must be true of the foci of a hyperbola?

The foci must lie on the transverse axis and be in the interior of the hyperbola.

If the transverse axis of a hyperbola is vertical, what do we know about the graph?

Where must the center of hyperbola be relative to its foci?

The center must be the midpoint of the line segment joining the foci.

## Algebraic

For the following exercises, determine whether the following equations represent hyperbolas. If so, write in standard form.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>3</mn><msup/></mrow></annotation-xml></semantics>[/itex] y 2 +2x=6

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 36 − y 2 9 =1

yes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 6 2 − y 2 3 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>5</mn><msup/></mrow></annotation-xml></semantics>[/itex] y 2 +4 x 2 =6x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>25</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −16 y 2 =400

yes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] x 2 4 2 − y 2 5 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +18x+ y 2 +4y−14=0

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 25 − y 2 36 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 5 2 − y 2 6 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5,0 ),( −5,0 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 61 ,0 ),( − 61 ,0 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 6 5 x,y=− 6 5 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 100 − y 2 9 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 4 − x 2 81 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 2 2 − x 2 9 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,2 ),( 0,−2 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0, 85 ),( 0,− 85 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 9 x,y=− 2 9 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] y 2 −4 x 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−1 ) 2 9 − ( y−2 ) 2 16 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−1 ) 2 3 2 − ( y−2 ) 2 4 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4,2 ),( −2,2 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 6,2 ),( −4,2 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 4 3 ( x−1 )+2,y=− 4 3 ( x−1 )+2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y−6 ) 2 36 − ( x+1 ) 2 16 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 ) 2 49 − ( y+7 ) 2 49 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 ) 2 7 2 − ( y+7 ) 2 7 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 9,−7 ),( −5,−7 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2+7 2 ,−7 ),( 2−7 2 ,−7 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>−</mo><mn>9</mn><mo>,</mo><mi>y</mi><mo>=</mo><mo>−</mo><mi>x</mi><mo>−</mo><mn>5</mn></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −8x−9 y 2 −72y+112=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −54x+9 y 2 −54y+81=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x+3 ) 2 3 2 − ( y−3 ) 2 3 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,3 ),( −6,3 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3+3 2 ,1 ),( −3−3 2 ,1 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>+</mo><mn>6</mn><mo>,</mo><mi>y</mi><mo>=</mo><mo>−</mo><mi>x</mi></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −24x−36 y 2 −360y+864=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +24x+16 y 2 −128y+156=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y−4 ) 2 2 2 − ( x−3 ) 2 4 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,6 ),( 3,2 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,4+2 5 ),( 3,4−2 5 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 2 ( x−3 )+4,y=− 1 2 ( x−3 )+4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +40x+25 y 2 −100y+100=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 +2x−100 y 2 −1000y+2401=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y+5 ) 2 7 2 − ( x+1 ) 2 70 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1,2 ),( −1,−12 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −1,−5+7 101 ),( −1,−5−7 101 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 10 ( x+1 )−5,y=− 1 10 ( x+1 )−5

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +72x+16 y 2 +16y+4=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +24x−25 y 2 +200y−464=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x+3 ) 2 5 2 − ( y−4 ) 2 2 2 =1; vertices:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 2,4 ),( −8,4 ); foci:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3+ 29 ,4 ),( −3− 29 ,4 ); asymptotes:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 5 ( x+3 )+4,y=− 2 5 ( x+3 )+4

For the following exercises, find the equations of the asymptotes for each hyperbola.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 3 2 − x 2 3 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−3 ) 2 5 2 − ( y+4 ) 2 2 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 5 ( x−3 )−4,y=− 2 5 ( x−3 )−4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y−3 ) 2 3 2 − ( x+5 ) 2 6 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>9</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −18x−16 y 2 +32y−151=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 3 4 ( x−1 )+1,y=− 3 4 ( x−1 )+1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>16</mn><msup/></mrow></annotation-xml></semantics>[/itex] y 2 +96y−4 x 2 +16x+112=0

## Graphical

For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 49 − y 2 16 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 64 − y 2 4 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 9 − x 2 25 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>81</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −9 y 2 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y+5 ) 2 9 − ( x−4 ) 2 25 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 ) 2 8 − ( y+3 ) 2 27 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] y−3 ) 2 9 − ( x−3 ) 2 9 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −8x+16 y 2 −32y−52=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 2 −8x−25 y 2 −100y−109=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +8x+4 y 2 −40y+88=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>64</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +128x−9 y 2 −72y−656=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>16</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +64x−4 y 2 −8y−4=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>100</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +1000x+ y 2 −10y−2575=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +16x−4 y 2 +16y+16=0

For the following exercises, given information about the graph of the hyperbola, find its equation.

Vertices at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,0 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] −3,0 ) and one focus at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 5,0 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 9 − y 2 16 =1

Vertices at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,6 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,−6 ) and one focus at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,−8 ).

Vertices at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 1,1 ) and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 11,1 ) and one focus at<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 12,1 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−6 ) 2 25 − ( y−1 ) 2 11 =1

Center:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,0 );vertex:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0,−13 );one focus:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 0, 313 ).

Center:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4,2 );vertex:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 9,2 );one focus:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 4+ 26 ,2 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−4 ) 2 25 − ( y−2 ) 2 1 =1

Center:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,5 ); vertex:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,11 ); one focus:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 3,5+2 10 ).

For the following exercises, given the graph of the hyperbola, find its equation.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 16 − x 2 25 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 9 − ( x+1 ) 2 9 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x+3 ) 2 25 − ( y+3 ) 2 25 =1

## Extensions

For the following exercises, express the equation for the hyperbola as two functions, with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]as a function of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 4 − y 2 9 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>y</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 9 − x 2 1 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=3 x 2 +1 ,y( x )=−3 x 2 +1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 ) 2 16 − ( y+3 ) 2 25 =1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −16x+ y 2 −2y−19=0

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x )=1+2 x 2 +4x+5 ,y( x )=1−2 x 2 +4x+5

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −24x− y 2 −4y+16=0

## Real-World Applications

For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph.

The hedge will follow the asymptotes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mtext> and </mtext><mi>y</mi><mo>=</mo><mo>−</mo><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and its closest distance to the center fountain is 5 yards.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 25 − y 2 25 =1

The hedge will follow the asymptotes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2</mn><mi>x</mi><mtext> and </mtext><mi>y</mi><mo>=</mo><mn>−2</mn><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics>[/itex] and its closest distance to the center fountain is 6 yards.

The hedge will follow the asymptotes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 2 x  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 2 x, and its closest distance to the center fountain is 10 yards.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 100 − y 2 25 =1

The hedge will follow the asymptotes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 3 x and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 2 3 x, and its closest distance to the center fountain is 12 yards.

The hedge will follow the asymptotes<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 3 4 x and y=− 3 4 x, and its closest distance to the center fountain is 20 yards.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 400 − y 2 225 =1

For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information.

The object enters along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>x</mi><mo>−</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>−</mo><mi>x</mi><mo>+</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]

The object enters along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2</mn><mi>x</mi><mo>−</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>−2</mn><mi>x</mi><mo>+</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−1 ) 2 0.25 − y 2 0.75 =1

The object enters along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>0.5</mn><mi>x</mi><mo>+</mo><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>−0.5</mn><mi>x</mi><mo>−</mo><mn>2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]

The object enters along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 3 x−1 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 3 x+1.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x−3 ) 2 4 − y 2 5 =1

The object It enters along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>3</mn><mi>x</mi><mo>−</mo><mn>9</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>−3</mn><mi>x</mi><mo>+</mo><mn>9.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]

## Glossary

center of a hyperbola
the midpoint of both the transverse and conjugate axes of a hyperbola
conjugate axis
the axis of a hyperbola that is perpendicular to the transverse axis and has the co-vertices as its endpoints
hyperbola
the set of all points<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) in a plane such that the difference of the distances between<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x,y ) and the foci is a positive constant
transverse axis
the axis of a hyperbola that includes the foci and has the vertices as its endpoints