20.5: Hyperbolas

Definitions and Theorems

Definition: Hyperbola

A hyperbola is the set of all points in a plane such that the difference of the distances between the point and two fixed points is a positive constant. Each fixed point is called a focus (plural: foci).

Definition: Transverse and Conjugate Axes, Vertex and Co-Vertex, and Center of a Hyperbola

The line segment that passes through the center of the hyperbola and has vertices as its endpoints is called transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of the hyperbola is the midpoint of both the transverse and conjugate axes.

Theorem: Standard Form of the Equation of a Hyperbola Centered at the Origin

The standard form of the equation of a hyperbola with center \((0,0)\) and transverse axis on the \(x\)-axis is\[\dfrac{x^2}{v_x^2}-\dfrac{y^2}{v_y^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• the length of the transverse axis is \(2 v_x\)
• the coordinates of the vertices are \(( \pm v_x, 0)\)
• the length of the conjugate axis is \(2 v_y\)
• the coordinates of the co-vertices are \((0, \pm v_y)\)
• the coordinates of the foci are \(( \pm c, 0)\), where \(c^2= \left| v_x^2 + v_y^2 \right| \)
• the equations of the asymptotes are \( y = \pm \frac{v_y}{v_x} x \).

The standard form of the equation of a hyperbola with center \((0,0)\) and transverse axis on the \(y\)-axis is\[ \dfrac{y^2}{v_y^2}-\dfrac{x^2}{v_x^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• the length of the transverse axis is \(2 v_y\)
• the coordinates of the vertices are \((0, \pm v_y)\)
• the length of the conjugate axis is \(2 v_x\)
• the coordinates of the co-vertices are \(( \pm v_x, 0)\)
• the coordinates of the foci are \((0, \pm c)\), where \(c^2 = \left| v_x^2 + v_y^2 \right| \)
• the equations of the asymptotes are \( y = \pm \frac{v_y}{v_x} x \).

Proof

To derive the equation of a hyperbola centered at the origin, we begin with the foci \( (−c,0) \) and \( (c,0) \). The hyperbola is the set of all points \( (x,y) \) such that the difference of the distances from \( (x,y) \) to the foci is constant, as shown in the figure below.

If \((v_x, 0)\) is a vertex of the hyperbola (on the transverse axis), the distance from \((-c, 0)\) to \((v_x, 0)\) is \(v_x-(-c)=v_x+c\). The distance from \((c, 0)\) to \((v_x, 0)\) is \(c - v_x\) (this is where the difference in the proof between the ellipse and the hyperbola begin to differ - notice from the figure above that \( c > v_x \)). The difference of the distances from the foci to the vertex is\[(v_x+c)-(c - v_x) = 2v_x. \nonumber \]If \((x, y)\) is a point on the hyperbola, then we can define the following variables:\[ \begin{array}{rcl} d_1 & = & \text { the distance from }(-c, 0) \text { to }(x, y) \\[6pt] d_2 & = & \text { the distance from }(c, 0) \text { to }(x, y)\\[6pt] \end{array} \nonumber \]By the definition of a hyperbola, \(d_1- d_2\) is constant for any point \((x, y)\) on the hyperbola. We know that the difference of these distances is \(2 v_x\) for the vertex \((v_x, 0)\). It follows that \(d_1-d_2=2 v_x\) for any point on the hyperbola. We will begin the derivation by applying the Distance Formula. The rest of the derivation is algebraic.\[ \begin{array}{rrclcl}
& d_1 - d_2 & = & 2v_x & & \\[6pt]
\implies & \sqrt{\left( x - (-c) \right)^2 + \left( y - 0 \right)^2} - \sqrt{\left( x - c \right)^2 + \left( y - 0 \right)^2} & = & 2v_x & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & \sqrt{\left( x + c \right)^2 + y^2} - \sqrt{\left( x - c \right)^2 + y^2} & = & 2v_x & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & \sqrt{\left( x + c \right)^2 + y^2} & = & 2v_x + \sqrt{\left( x - c \right)^2 + y^2} & \quad & \left( \text{isolating a radical} \right) \\[6pt]
\implies & \left( x + c \right)^2 + y^2 & = & \left[ 2v_x + \sqrt{\left( x - c \right)^2 + y^2} \right]^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & \left( x + c \right)^2 + y^2 & = & 4v_x^2 + 4v_x \sqrt{\left( x - c \right)^2 + y^2} + \left( x - c \right)^2 + y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & x^2 + 2 c x + c^2 + y^2 & = & 4v_x^2 + 4v_x \sqrt{\left( x - c \right)^2 + y^2} + x^2 - 2c x + c^2 + y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 2 c x & = & 4v_x^2 + 4v_x \sqrt{\left( x - c \right)^2 + y^2} - 2c x & \quad & \left( \text{subtracting }x^2, \, c^2, \text{ and }y^2 \right. \\[6pt]
 &  &  & & \quad & \left. \text{ from both sides} \right) \\[6pt]
\implies & 4 c x - 4v_x^2 & = & 4v_x \sqrt{\left( x - c \right)^2 + y^2} & \quad & \left( \text{isolating the radical} \right) \\[6pt]
\implies & c x - v_x^2 & = & v_x \sqrt{\left( x - c \right)^2 + y^2} & \quad & \left( \text{dividing both sides by }4 \right) \\[6pt]
\implies & \left(c x - v_x^2\right)^2 & = & v_x^2 \left(\left( x - c \right)^2 + y^2\right) & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & c^2 x^2 - 2v_x^2 c x + v_x^4 & = & v_x^2 \left(x^2 - 2cx + c^2 + y^2\right) & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & c^2 x^2 - 2v_x^2 c x + v_x^4 & = & v_x^2 x^2 - 2 v_x^2 cx + v_x^2 c^2 + v_x^2 y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & c^2 x^2 + v_x^4 & = & v_x^2 x^2 + v_x^2 c^2 + v_x^2 y^2 & \quad & \left( \text{adding }2v_x^2 c x\text{ to both sides} \right) \\[6pt]
\implies & v_x^4 - v_x^2 c^2 & = & v_x^2 x^2 - c^2 x^2 + v_x^2 y^2 & \quad & \left( \text{subtracting }v_x^2 c^2 \text{ and }c^2 x^2 \right. \\[6pt]
 &  &  & & \quad & \left. \text{ from both sides} \right) \\[6pt]
\implies & v_x^2(v_x^2 - c^2) & = & x^2(v_x^2 - c^2 ) + v_x^2 y^2 & \quad & \left( \text{factor} \right) \\[6pt]
\implies & -v_x^2(v_x^2 - c^2) & = & -x^2(v_x^2 - c^2 ) - v_x^2 y^2 & \quad & \left( \text{multiplying both sides by }-1 \right) \\[6pt]
\implies & v_x^2(c^2 - v_x^2) & = & x^2(c^2 - v_x^2) - v_x^2 y^2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & v_x^2 b^2 & = & x^2 b^2 - v_x^2 y^2 & \quad & \left( \text{let }b^2 = c^2 - v_x^2 \right) \\[6pt]
\implies & \dfrac{v_x^2 b^2}{v_x^2 b^2} & = & \dfrac{x^2 b^2}{v_x^2 b^2} - \dfrac{v_x^2 y^2}{v_x^2 b^2} & \quad & \left( \text{divide both sides by }v_x^2 b^2 \right) \\[6pt]
\implies & 1 & = & \dfrac{x^2}{v_x^2} - \dfrac{y^2}{b^2} & \quad & \left( \text{simplify} \right) \\[6pt]
\end{array} \nonumber \]At this point in the proof of the equation for the ellipse, we were able to let \( x = 0 \) to determine the meaning of \( b \); however, the hyperbola opening horizontally in the figure above does not exist at \( x = 0 \). Therefore, such a simple maneuver is not possible here. Instead, let's consider our result,\[ \dfrac{x^2}{v_x^2} - \dfrac{y^2}{b^2} = 1, \nonumber \]and solve for \( y \).\[ \begin{array}{rrclcl}
& \dfrac{x^2}{v_x^2} - \dfrac{y^2}{b^2} & = & 1 & & \\[6pt]
\implies & \dfrac{x^2}{v_x^2} & = & 1 + \dfrac{y^2}{b^2} & \quad & \left( \text{adding }\frac{y^2}{b^2}\text{ to both sides} \right) \\[6pt]
\implies & \dfrac{x^2}{v_x^2} - 1 & = & \dfrac{y^2}{b^2} & \quad & \left( \text{subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & \pm \sqrt{\dfrac{x^2}{v_x^2} - 1} & = & \dfrac{y}{b} & \quad & \left( \text{taking the square root of both sides} \right) \\[6pt]
\implies & \pm b \sqrt{\dfrac{x^2}{v_x^2} - 1} & = & y & \quad & \left( \text{multiplying both sides by }b \right) \\[6pt]
\implies & \pm b \sqrt{\dfrac{x^2}{v_x^2}\left(1 - \dfrac{v_x^2}{x^2}\right)} & = & y & \quad & \left( \text{factoring }\dfrac{x^2}{v_x^2}\text{ within the radicand} \right) \\[6pt]
\implies & \pm b \left| \dfrac{x}{v_x} \right| \sqrt{1 - \dfrac{v_x^2}{x^2}} & = & y & \quad & \left( \text{simplifying using }\sqrt{\blacksquare^2}= |\blacksquare| \right) \\[6pt]
\end{array} \nonumber \]Since we are requiring \( v_x > 0 \), \( |v_x| = v_x \). Moreover, for simplicity (and without loss of generality), let's only consider points on the right branch of the hyperbola. That is, let's assume \( x > 0 \). Then our final result becomes\[ y = \pm \dfrac{b}{v_x} x \sqrt{1 - \dfrac{v_x^2}{x^2}}. \nonumber \]Using the same logic we developed when discussing slant asymptotes (see the section on graphing rational functions), we consider what happens as \( x \to \infty \). To truly grasp this concept, you must keep in mind that \( v_x \) is a constant.\[ \begin{array}{rcl}
\text{As } x \to \infty, & \quad & \dfrac{v_x^2}{x^2} \to 0 \\[6pt]
\text{As } \dfrac{v_x^2}{x^2} \to 0, & \quad & 1 - \dfrac{v_x^2}{x^2} \to 1 \\[6pt]
\text{As } 1 - \dfrac{v_x^2}{x^2} \to 1, & \quad & \sqrt{1 - \dfrac{v_x^2}{x^2}} \to 1 \\[6pt]
\end{array} \nonumber \]Thus, as \( x \to \infty \), the \( y \)-values for the points on our hyperbola approach\[ y = \pm \dfrac{b}{v_x} x \sqrt{1 - \dfrac{v_x^2}{x^2}} \to \pm \dfrac{b}{v_x} x \cdot 1 = \pm \frac{b}{v_x} x. \nonumber \]That is, the lines \( y = \frac{b}{v_x} x \) and \( y = -\frac{b}{v_x} x \) are slant asymptotes (guidelines for graphing) as \( x \to \infty \). Both of these are linear equations where the slopes are \( \pm \frac{b}{v_x} \). Since slope is "rise over run," or "increase in \( y \) over increase in \( x \)," we can interpret the numerator of these slopes as the corresponding change in \( y \) when \( x \)-values change by \( v_x \). Making the notational change, \( b = v_y \), we arrive at our desired result: the equation of the hyperbola centered at the origin and opening horizontally is\[ \dfrac{x^2}{v_x^2} - \dfrac{y^2}{v_y^2} = 1, \nonumber \]where the slant asymptotes are \( y = \pm \frac{v_y}{v_x} x \).

The proof for the equation of the hyperbola opening vertically is similar.

Theorem: Standard Form of the Equation of a Hyperbola Centered Off-Origin

The standard form of the equation of a hyperbola with center \((h, k)\) and transverse axis parallel to the \(x\)-axis is\[\dfrac{(x-h)^2}{v_x^2}-\dfrac{(y-k)^2}{v_y^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• the length of the transverse axis is \(2 v_x\)
• the coordinates of the vertices are \((h \pm v_x, k)\)
• the length of the conjugate axis is \(2 v_y\)
• the coordinates of the co-vertices are \((h, k \pm v_y)\)
• the coordinates of the foci are \((h \pm c, k)\), where \(c^2 = \left| v_x^2 + v_y^2 \right| \)
• the equations of the asymptotes are \( y - k = \pm \frac{v_y}{v_x} (x - h) \).

The standard form of the equation of a hyperbola with center \((h, k)\) and transverse axis parallel to the \(y\)-axis is\[ \dfrac{(y-k)^2}{v_y^2} - \dfrac{(x-h)^2}{v_x^2}=1, \nonumber \]where

• \( v_x \) and \( v_y \) are positive
• the length of the transverse axis is \(2 v_y\)
• the coordinates of the vertices are \((h, k \pm v_y)\)
• the length of the conjugate axis is \(2 v_x\)
• the coordinates of the co-vertices are \((h \pm v_x, k)\)
• the coordinates of the foci are \((h, k \pm c)\), where \(c^2 = \left| v_x^2 + v_y^2 \right| \)
• the equations of the asymptotes are \( y - k = \pm \frac{v_y}{v_x} (x - h) \).