9.2: Hyperbolas
In the last section, we learned that planets have approximately elliptical orbits around the sun. When an object like a comet is m oving quickly, it is able to escape the gravitational pull of the sun and follows a path with the shape of a hyperbola . Hyperbolas are curves that can help us find the location of a ship, describe the shape of cooling towers, or calibrate seismological equipment.
The hyperbola is another type of conic section created by intersecting a plane with a double cone, as shown below (Pbroks13 (commons.wikimedia.org/wiki/F...with_plane.svg), “Conic sections with plane”, cropped to show only a hyperbola by L Michaels, CC BY 3.0).
The word “hyperbola” derives from a Greek word meaning “excess.” The English word “hyperbole” means exaggeration. We can think of a hyperbola as an excessive or exaggerated ellipse, one turned inside out.
We defined an ellipse as the set of all points where the sum of the distances from that point to two fixed points is a constant. A hyperbola is the set of all points where the absolute value of the difference of the distances from the point to two fixed points is a constant.
Definition: Hyperbola
A hyperbola is the set of all points \(Q\;\left( x,y \right)\) for which the absolute value of the difference of the distances to two fixed points \(F_1\left( x_1,y_1 \right)\) and \(F_2\left( x_2,y_2 \right)\) called the foci (plural for focus) is a constant \(k\): \[\left| {d\left( Q,F_1 \right) - d\left( Q,F_2 \right)} \right| = k\]
- The transverse axis is the line passing through the foci.
- Vertices are the points on the hyperbola which intersect the transverse axis.
- The transverse axis length is the length of the line segment between the vertices.
- The center is the midpoint between the vertices (or the midpoint between the foci).
- The other axis of symmetry through the center is the conjugate axis .
- The two disjoint pieces of the curve are called branches .
- A hyperbola has two asymptotes .
Which axis is the transverse axis will depend on the orientation of the hyperbola. As a helpful tool for graphing hyperbolas, it is common to draw a central rectangle as a guide. This is a rectangle drawn around the center with sides parallel to the coordinate axes that pass through each vertex and co-vertex. The asymptotes will follow the diagonals of this rectangle.
Hyperbolas Centered at the Origin
From the definition above we can find an equation of a hyperbola. We will find it for a hyperbola centered at the origin \(C\left( 0,0 \right)\) opening horizontally with foci at
\[F_1\left( c,0 \right)\text{ and }F_2\left( - c,0 \right)\text{ where }c > 0\nonumber\]
Suppose \(Q\left( x,y \right)\) is a point on the hyperbola. The distances from \(Q\) to \(F_1\) and \(Q\) to \(F_2\) are:
\[d\left( Q, F_1 \right) = \sqrt {\left( x - c \right)^2 + \left( y - 0 \right)^2} = \sqrt {\left( x - c \right)^2 + y^2} \nonumber\]
\[d\left( Q,F_2 \right) = \sqrt {\left( x - \left( - c \right) \right)^2 + \left( y - 0 \right)^2} = \sqrt {\left( x + c \right)^2 + y^2} \nonumber\]
From the definition, the absolute value of the difference should be constant:
\[\left| d\left( Q, F_1 \right) - d\left(Q, F_2 \right) \right| = \left| \sqrt {\left( x - c \right)^2 + y^2} - \sqrt {\left( x + c \right)^2 + y^2 } \right| = k\nonumber\]
Substituting in one of the vertices \(\left( a,0 \right)\), we can determine \(k\) in terms of \(a\):
\[\left| \sqrt {\left( a - c \right)^2 + 0^2} - \sqrt {\left( a + c \right)^2 + 0^2 } \right| = k\nonumber\]
\[\left| \left| a - c \right| - \left| a + c \right| \right| = k\nonumber\] Since \(c > a\), \(\left| a - c \right| = c - a\)
\[\left| (c - a) - (a + c) \right| = k\nonumber\]
\[k = \left| - 2a \right| = \left| 2a \right|\nonumber\]
Using \(k = 2a\) and removing the absolute values,
\[\sqrt {\left( x - c \right)^2 + y^2} - \sqrt {\left( x + c \right)^2+ y^2} = \pm 2a\nonumber\]Move one radical
\[\sqrt {\left( x - c \right)^2 + y^2} = \pm 2a + \sqrt {\left( x + c \right)^2 + y^2} \nonumber\]Square both sides
\[\left( x - c \right)^2 + y^2 = 4a^2 \pm 4a\sqrt {\left( x + c \right)^2 + y^2} + \left( x + c \right)^2 + y^2\nonumber\]Expand
\[x^2 - 2xc + c^2 + y^2 = 4a^2 \pm 4a\sqrt {\left( x + c \right)^2 + y^2} + x^2 + 2xc + c^2 + y^2\nonumber\]
Combining like terms leaves
\[ - 4xc = 4a^2 \pm 4a\sqrt {\left( x + c \right)^2 + y^2} \nonumber\]Divide by 4
\[ - xc = a^2 \pm a \sqrt {\left( x + c \right)^2 + y^2} \nonumber\]Isolate the radical
\[ \pm a \sqrt {\left( x + c \right)^2 + y^2} = - a^2 - xc\nonumber\]Square both sides again
\[a^2\left( \left( x + c \right)^2 + y^2 \right) = a^4 + 2a^2xc + x^2c^2\nonumber\]Expand and distribute
\[a^2x^2 + 2a^2xc + a^2c^2 + a^2y^2 = a^4 + 2a^2xc + x^2c^2\nonumber\]Combine like terms
\[a^2y^2 + a^2c^2 - a^4 = x^2c^2 - a^2x^2\nonumber\]Factor common terms
\[a^2y^2 + a^2\left( c^2 - a^2 \right) = \left( c^2 - a^2 \right)x^2\nonumber\]
Let \(b^2 = c^2 - a^2\). Since \(c > a\), \(b > 0\). Substituting \(b^2\) for \(c^2 - a^2\) leaves
\[a^2y^2 +a^2b^2 = b^2x^2\nonumber\]Divide both sides by \(a^2b^2\)
\[\dfrac{y^2}{b^2} + 1 = \dfrac{x^2}{a^2}\nonumber\]Rewrite
\[\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\nonumber\]
We can see from the graphs of the hyperbolas that the branches appear to approach asymptotes as \(x\) gets large in the negative or positive direction. The equations of the horizontal hyperbola asymptotes can be derived from its standard equation.
\[\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\nonumber\]Solve for \(y\)
\[y^2 = b^2\left( \dfrac{x^2}{a^2} - 1 \right)\nonumber\]Rewrite 1 as \(\dfrac{x^2}{a^2}\dfrac{a^2}{x^2}\)
\[y^2 = b^2\left( \dfrac{x^2}{a^2} - \dfrac{x^2}{a^2}\dfrac{a^2}{x^2} \right)\nonumber\]Factor out \(\dfrac{x^2}{a^2}\)
\[y^2 = b^2\dfrac{x^2}{a^2}\left( 1 - \dfrac{a^2}{x^2} \right)\nonumber\]Take the square root
\[y = \pm \dfrac{b}{a}x\sqrt {1 - \dfrac{a^2}{x^2}} \nonumber\]
As \(x \to \pm \infty\) the quantity \(\dfrac{a^2}{x^2} \to 0\) and \(\sqrt {1 - \dfrac{a^2}{x^2}} \to 1\), so the asymptotes are \(y = \pm \dfrac{b}{a}x\).
Similarly, for vertical hyperbolas the asymptotes are \(y = \pm \dfrac{a}{b}x\).
The standard form of an equation of a hyperbola centered at the origin C\(\left( {0,0} \right)\) depends on whether it opens horizontally or vertically. The following table gives the standard equation, vertices, foci, asymptotes, construction rectangle vertices, and graph for each.
| Opens | Horizontally | Vertically |
|---|---|---|
| Standard Equation | \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) | \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\) |
| Vertices | \((-a, 0)\) and \((a, 0)\) | \((0, -a)\) and \((0, a)\) |
| Foci |
\((-c, 0)\) and \((c, 0)\)
where \(b^2 = c^2 - a^2\) |
\((0, -c)\) and \((0, c)\) |
| Asymptotes | \(y = \pm \dfrac{b}{a} x\) | \(y = \pm \dfrac{a}{b} x\) |
| Construction Rectangle Vertices | \((a, b)\), \((-a, b)\), \((a, -b)\), \((-a, -b)\) | \((b, a)\), \((-b, a)\), \((b, -a)\), \((-b, -a)\) |
| Graph |
Example \(\PageIndex{1}\)
Put the equation of the hyperbola \(y^2 - 4x^2 = 4\) in standard form. Find the vertices, length of the transverse axis, and the equations of the asymptotes. Sketch the graph. Check using a graphing utility.
Solution
The equation can be put in standard form \(\dfrac{y^2}{4} - \dfrac{x^2}{1} = 1\) by dividing by 4.
Comparing to the general standard equation \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\) we see that \(a = \sqrt 4 = 2\) and \(b = \sqrt 1 = 1\).
Since the \(x\) term is subtracted, the hyperbola opens vertically and the vertices lie on the \(y\)-axis at (0, \(\pm a\)) = (0, \(\pm\)2).
The length of the transverse axis is \[2\left( a \right) = 2\left( 2 \right) = 4\nonumber\]
Equations of the asymptotes are \[y = \pm \dfrac{a}{b}x\text{ or } = \pm 2x\nonumber\]
To sketch the graph we plot the vertices of the construction rectangle at \((\pm b, \pm a)\) or (-1,-2), (-1,2), (1,-2), and (1,2). The asymptotes are drawn through the diagonals of the rectangle and the vertices plotted. Then we sketch in the hyperbola, rounded at the vertices and approaching the asymptotes.
To check on a graphing utility, we must solve the equation for \(y\). Isolating \(y^2\) gives us\(y^2 = 4\left( 1 + x^2 \right)\).
Taking the square root of both sides we find \[y = \pm 2\sqrt {1 + x^2} \nonumber\]
Under Y= enter the two halves of the hyperbola and the two asymptotes as \(y = 2\sqrt {1 + x^2} \), \(y = - 2\sqrt {1 + x^2} \), \(y = 2x\), and \(y = - 2x\). Set the window to a comparable scale to the sketch with xmin = -4, xmax = 4, ymin= -3, and ymax = 3.
Sometimes we are given the equation. Sometimes we need to find the equation from a graph or other information.
Example \(\PageIndex{2}\)
Find the standard form of the equation for a hyperbola with vertices at (-6,0) and (6,0) and asymptote \(y = \dfrac{4}{3}x\).
Solution
Since the vertices lie on the \(x\)-axis with a midpoint at the origin, the hyperbola is horizontal with an equation of the form \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). The value of a is the distance from the center to a vertex. The distance from (6, 0) to (0, 0) is 6, so \(a\) = 6.
The asymptotes follow the form \(y = \pm \dfrac{b}{a}x\). From \(y = \dfrac{3}{4}x\) we see \(\dfrac{3}{4} = \dfrac{b}{a}\) and substituting \(a = 6\) give us \(\dfrac{3}{4} = \dfrac{b}{6}\). Solving yields \(b = 8\).
The equation of the hyperbola in standard form is \[\dfrac{x^2}{6^2} - \dfrac{y^2}{8^2} = 1\text{ or }\dfrac{x^2}{36} - \dfrac{y^2}{64} = 1\nonumber\]
Exercise \(\PageIndex{2}\)
Find the standard form of the equation for a hyperbola with vertices at (0, -8) and (0, 8) and asymptote \(y = 2x\)
- Answer
-
The vertices are on the \(y\) axis so this is a vertical hyperbola.
The center is at the origin.
\[a = 8\nonumber\]Using the asymptote slope, \(\dfrac{8}{b} = 2\), so \(b = 4\).
\[\dfrac{y^2}{64} - \dfrac{x^2}{16} = 1\nonumber\]
Example \(\PageIndex{3}\)
Find the standard form of the equation for a hyperbola with vertices at (0, 9) and (0, -9) and passing through the point (8, 15).
Solution
Since the vertices lie on the \(y\)-axis with a midpoint at the origin, the hyperbola is vertical with an equation of the form \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\). The value of a is the distance from the center to a vertex. The distance from (0, 9) to (0, 0) is 9, so \(a = 9\).
Substituting \(a = 9\) and the point (8, 15) gives \(\dfrac{15^2}{9^2} - \dfrac{8^2}{b^2} = 1\). Solving for \(b\) yields
\[b = \sqrt {\dfrac{9^2 \left( 8^2 \right)}{15^2 - 9^2}} = 6\nonumber\]
The standard equation for the hyperbola is \[\dfrac{y^2}{9^2} - \dfrac{x^2}{6^2} = 1\text{ or}\dfrac{y^2}{81} - \dfrac{x^2}{36} = 1\nonumber\]
Hyperbolas Not Centered at the Origin
Not all hyperbolas are centered at the origin. The standard equation for one centered at (\(h\), \(k\)) is slightly different.
IN STANDARD FORM
The standard form of an equation of a hyperbola centered at \(C\left( h,k \right)\) depends on whether it opens horizontally or vertically. The table below gives the standard equation, vertices, foci, asymptotes, construction rectangle vertices, and graph for each.
| Opens | Horizontally | Vertically |
| Standard Equation | \(\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1\) | \(\dfrac{(y - k)^2}{a^2} - \dfrac{(x - h)^2}{b^2} = 1\) |
| Vertices | \((h \pm a, k)\) | \((h, k \pm a)\) |
| Foci |
\((h \pm c, k)\)
where \(b^2 = c^2 - a^2\) |
\((h, k \pm c)\)
where \(b^2 = c^2 - a^2\) |
| Asymptotes | \(y - k = \pm \dfrac{b}{a} (x - h)\) | \(y - k = \pm \dfrac{a}{b} (x - h)\) |
| Construction Rectangle Vertices | \((h \pm a, k \pm b)\) | \((h \pm b, k \pm a)\) |
| Graph |
Example \(\PageIndex{4}\)
Write an equation for the hyperbola in the graph shown.
Solution
The center is at (2, 3), where the asymptotes cross. It opens vertically, so the equation will look like \(\dfrac{\left( y - 3 \right)^2}{a^2} - \dfrac{\left( x - 2 \right)^2}{b^2} = 1\).
The vertices are at (2, 2) and (2, 4). The distance from the center to a vertex is \(a = 4 - 3 = 1\).
If we were to draw in the construction rectangle, it would extend from \(x = -1\) to \(x = 5\). The distance from the center to the right side of the rectangle gives \(b = 5 - 2 = 3\).
The standard equation of this hyperbola is \[\dfrac{\left( y - 3 \right)^2}{1^2} - \dfrac{\left( x - 2 \right)^2}{3^2} = 1\text{, or }\left( y - 3 \right)^2 - \dfrac{\left( x - 2\right)^2}{9} = 1\nonumber\]
Example \(\PageIndex{5}\)
Put the equation of the hyperbola \(9x^2 + 18x - 4y^2 + 16y = 43\) in standard form. Find the center, vertices, length of the transverse axis, and the equations of the asymptotes. Sketch the graph, then check on a graphing utility.
Solution
To rewrite the equation, we complete the square for both variables to get
\(9\left( x^2 + 2x + 1 \right) - 4\left( y^2 - 4y + 4 \right) = 43 + 9 - 16\)
\(9\left( x + 1 \right)^2 - 4\left( y - 2 \right)^2 = 36\)
Dividing by 36 gives the standard form of the equation, \[\dfrac{\left( x + 1 \right)^2}{4} - \dfrac{\left( y - 2 \right)^2}{9} = 1\nonumber\]
Comparing to the general standard equation \(\dfrac{\left( x - h \right)^2}{a^2} - \dfrac{\left( h - k \right)^2}{b^2} = 1\) we see that \[a = \sqrt 4 = 2\text{and }b = \sqrt 9 = 3\nonumber\]
Since the \(y\) term is subtracted, the hyperbola opens horizontally.
The center is at (\(h\), \(k\)) = (-1, 2).
The vertices are at (\(h \pm a\), \(k\)) or (-3, 2) and (1,2).
The length of the transverse axis is \(2\left( a \right) = 2\left( 2 \right) = 4\).
Equations of the asymptotes are \[y - k = \pm \dfrac{b}{a}\left( x - h \right)\text{ or }y - 2 = \pm \dfrac{3}{2}\left( x + 1 \right)\nonumber\]
To sketch the graph we plot the corners of the construction rectangle at (\(h \pm a\), \(k \pm b\)) or (1, 5), (1, -1), (-3, 5), and (-3, -1). The asymptotes are drawn through the diagonals of the rectangle and the vertices plotted. Then we sketch in the hyperbola rounded at the vertices and approaching the asymptotes.
To check on a graphing utility, we must solve the equation for \(y\).
\[y = 2 \pm \sqrt {9\left( \dfrac{\left( x + 1 \right)^2}{4} - 1 \right)} \nonumber\]
Under Y= enter the two halves of the hyperbola and the two asymptotes as \[y = 2 + \sqrt {9\left( \dfrac{\left( x + 1 \right)^2}{4} - 1 \right)} \quad y = 2 - \sqrt {9\left( \dfrac{\left( x + 1 \right)^2}{4} - 1 \right)} \quad y = \dfrac{3}{2}\left( x + 1 \right) + 2\quad\text{, and }y = - \dfrac{3}{2}\left( x + 1 \right) + 2\nonumber\] Set the window to a comparable scale to the sketch, then graph.
Note that the gaps you see on the calculator are not really there; they’re a limitation of the technology.
Example \(\PageIndex{6}\)
Find the standard form of the equation for a hyperbola with vertices at \(( - 2, - 5)\) and \(( - 2, 7)\), and asymptote \(y = \dfrac{3}{2}x + 4\).
Solution
Since the vertices differ in the \(y\)-coordinates, the hyperbola opens vertically with an equation of the form \(\dfrac{\left( y - k \right)^2}{a^2} - \dfrac{\left( x - h\right)^2}{b^2} = 1\) and asymptote equations of the form \[y - k = \pm \dfrac{a}{b}\left( x - h \right)\nonumber\]
The center will be halfway between the vertices, at \(\left( - 2,\dfrac{ - 5 + 7}{2} \right) = ( - 2,1)\).
The value of a is the distance from the center to a vertex. The distance from \(( - 2,1)\) to \(( - 2, - 5)\) is 6, so \(a = 6\).
While our asymptote is not given in the form \(y - k = \pm \dfrac{a}{b}\left( {x - h} \right)\), notice this equation would have slope \(\dfrac{a}{b}\). We can compare that to the slope of the given asymptote equation to find b. Setting \(\dfrac{3}{2} = \dfrac{a}{b}\) and substituting \(a = 6\) gives us \(b = 4\).
The equation of the hyperbola in standard form is \[\dfrac{\left( y - 1 \right)^2}{6^2} - \dfrac{\left( x + 2 \right)^2}{4^2} = 1\text{ or }\dfrac{\left( y - 1 \right)^2}{36} - \dfrac{\left( x + 2 \right)^2}{16} = 1\nonumber\]
Exercise \(\PageIndex{2}\)
Find the center, vertices, length of the transverse axis, and equations of the asymptotes for the hyperbola \(\dfrac{\left( x + 5 \right)^2}{9} - \dfrac{\left( y - 2 \right)^2}{36} = 1\).
- Answer
-
Center (-5, 2). This is a horizontal hyperbola. \(a = 3\). \(b = 6\).
transverse axis length 6,
Vertices will be at (-5 \(\pm\) 3,2) = (-2, 2) and (-8, 2),
Asymptote slope will be \(\dfrac{6}{3} = 2\). Asymptotes: \(y - 2 = \pm 2\left( x + 5 \right)\)
Hyperbola Foci
The location of the foci can play a key role in hyperbola application problems. To find them, we need to find the length from the center to the foci, \(c\), using the equation \(b^2 = c^2 - a^2\). It looks similar to, but is not the same as, the Pythagorean Theorem.
Compare this with the equation to find length \(c\) for ellipses, which is \(b^2 = a^2 - c^2\). If you remember that for the foci to be inside the ellipse they have to come before the vertices \((c < a)\), it’s clear why we would calculate \(a^2\) minus \(c^2\). To be inside a hyperbola, the foci have to go beyond the vertices \((c > a)\), so we can see for hyperbolas we need \(c^2\) minus \(a^2\), the opposite.
Example\(\PageIndex{7}\)
Find the foci of the hyperbola \(\dfrac{\left( y + 1 \right)^2}{4} - \dfrac{\left( x - 3 \right)^2}{5} = 1\).
Solution
The hyperbola is vertical with an equation of the form \[\dfrac{\left( y - k \right)^2}{a^2} - \dfrac{\left( x - h \right)^2}{b^2} = 1\nonumber\]
The center is at (\(h\), \(k\)) = (3, -1). The foci are at (\(h\), \(k \pm c\)).
To find length c we use \(b^2 = c^2 - a^2\). Substituting gives \[5 = c^2 - 4\text{ or }c = \sqrt 9 = 3\nonumber\]
The hyperbola has foci (3, -4) and (3, 2).
Example \(\PageIndex{8}\)
Find the standard form of the equation for a hyperbola with foci (5, -8) and (-3, -8) and vertices (4, -8) and (-2, -8).
Solution
Since the vertices differ in the \(x\)-coordinates, the hyperbola opens horizontally with an equation of the form \(\dfrac{\left( x - h \right)^2}{a^2} - \dfrac{\left( y - k \right)^2}{b^2} = 1\).
The center is at the midpoint of the vertices \[\left( \dfrac{x_1 + x_2}{2},\dfrac{y_1 + y_2}{2} \right) = \left( \dfrac{4 + \left( - 2 \right)}{2},\dfrac{ - 8 + \left( - 8 \right)}{2} \right) = \left( 1, - 8 \right)\nonumber\]
The value of \(a\) is the horizontal length from the center to a vertex, or \[a = 4 - 1 = 3\nonumber\]
The value of \(c\) is the horizontal length from the center to a focus, or \[= 5 - 1 = 4\nonumber\]
To find length \(b\) we use \(b^2 = c^2 - a^2\). Substituting gives \[b^2 = 16 - 9 = 7\nonumber\]
The equation of the hyperbola in standard form is \[\dfrac{\left( x - 1 \right)^2}{3^2} - \dfrac{\left( y - \left( - 8 \right) \right)^2}{7} = 1\text{ or }\dfrac{\left( x - 1\right)^2}{9} - \dfrac{\left( y + 8\right)^2}{7} = 1\nonumber\]
Exercise \(\PageIndex{3}\)
Find the standard form of the equation for a hyperbola with focus (1, 9), vertex (1, 8), center (1, 4).
- Answer
-
Focus, vertex, and center have the same \(x\) value so this is a vertical hyperbola.
Using the vertex and center, \[a = 9 – 4 = 5\nonumber\]
Using the focus and center, \[c = 8 – 4 = 4\nonumber\]
\[b^2 =5^2 - 4^2\quad b = 3\nonumber\]
\[\dfrac{\left( y - 4 \right)^2}{16} - \dfrac{\left( x - 1 \right)^2}{9} = 1\nonumber\]
LORAN
Before GPS, the Long Range Navigation (LORAN) system was used to determine a ship’s location. Two radio stations A and B simultaneously sent out a signal to a ship. The difference in time it took to receive the signal was computed as a distance locating the ship on the hyperbola with the A and B radio stations as the foci. A second pair of radio stations C and D sent simultaneous signals to the ship and computed its location on the hyperbola with C and D as the foci. The point P where the two hyperbolas intersected gave the location of the ship.
Example \(\PageIndex{9}\)
Stations A and B are 150 kilometers apart and send a simultaneous radio signal to the ship. The signal from B arrives 0.0003 seconds before the signal from A. If the signal travels 300,000 kilometers per second, find the equation of the hyperbola on which the ship is positioned.
Solution
Stations A and B are at the foci, so the distance from the center to one focus is half the distance between them, giving \(c = \dfrac{1}{2}(150) = 75\) km.
By letting the center of the hyperbola be at (0, 0) and placing the foci at (\(\pm\)75,0), the equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) for a hyperbola centered at the origin can be used.
The difference of the distances of the ship from the two stations is \(k = 300,000\dfrac{\rm{km}}{\rm{s}} \cdot (0.0003 \rm{s}) = 90 \rm{km}\). From our derivation of the hyperbola equation we determined \(k = 2a\), so \[a = \dfrac{1}{2}(90) = 45\nonumber\]
Substituting \(a\) and \(c\) into \(b^2 = c^2 - a^2\) yields \[b^2 = 75^2 - 45^2 = 3600\nonumber\]
The equation of the hyperbola in standard form is \[\dfrac{x^2}{45^2} - \dfrac{y^2}{3600} = 1\text{ or }\dfrac{x^2}{2025} - \dfrac{y^2}{3600} = 1\nonumber\]
To determine the position of a ship using LORAN, we would need an equation for the second hyperbola and would solve for the intersection. We will explore how to do that in the next section.
Important Topics of This Section
- Hyperbola Definition
- Hyperbola Equations in Standard Form
- Hyperbola Foci
- Applications of Hyperbolas
- Intersections of Hyperbolas and Other Curves