MA001: College Algebra

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 Figure 2.

Figure 2 A hyperbola

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 $(x,y)$ in a plane such that the difference of the distances between $(x,y)$ 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. The conjugate 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 Figure 3.

Figure 3 Key features of the hyperbola

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 a Hyperbola Centered at the Origin

Let $(−c,0)$ and $(c,0)$ be the foci of a hyperbola centered at the origin. 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. See Figure 4.

Figure 4

If $(a, 0)$ is a vertex of the hyperbola, the distance from $(-c, 0)$ to $(a, 0)$ is $a-(-c)=a+c$. The distance from $(c, 0)$ to $(a, 0)$ is $c-a$. The difference of the distances from the foci to the vertex is

$(a+c)-(c-a)=2 a$

If $(x, y)$ is a point on the hyperbola, we can define the following variables:

$\begin{aligned} &d_{2}=\text { the distance from }(-c, 0) \text { to }(x, y) \\ &d_{1}=\text { the distance from }(c, 0) \text { to }(x, y) \end{aligned}$

By definition of a hyperbola, $d_{2}-d_{1}$ is constant for any point $(x, y)$ on the hyperbola. We know that the difference of these distances is $2 a$ for the vertex $(a, 0)$. It follows that $d_{2}-d_{1}=2 a$ 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.

$\begin{array}{ll} d_{2}-d_{1}=\sqrt{(x-(-c))^{2}+(y-0)^{2}}+\sqrt{(x-c)^{2}+(y-0)^{2}}=2 a & \text { Distance formula } \\ \sqrt{(x+c)^{2}+y^{2}}-\sqrt{(x-c)^{2}+y^{2}}=2 a & \text { Simplify expressions. } \\ \sqrt{(x+c)^{2}+y^{2}}=2 a+\sqrt{(x-c)^{2}+y^{2}} & \text { Move radical to opposite side. } \\ (x+c)^{2}+y^{2}=\left(2 a+\sqrt{(x-c)^{2}+y^{2}}\right)^{2} & \text { Square both sides. } \\ x^{2}+2 c x+c^{2}+y^{2}=4 a^{2}+4 a \sqrt{(x-c)^{2}+y^{2}}+(x-c)^{2}+y^{2} & \text { Expand the squares. } \\ x^{2}+2 c x+c^{2}+y^{2}=4 a^{2}+4 a \sqrt{(x-c)^{2}+y^{2}}+x^{2}-2 c x+c^{2}+y^{2} & \text { Expand remaining squares. } \\ 2 c x=4 a^{2}+4 a \sqrt{(x-c)^{2}+y^{2}}-2 c x & \text { Combine like terms. } \\ 4 c x-4 a^{2}=4 a \sqrt{(x-c)^{2}+y^{2}} & \text { Isolate the radical. } \\ c x-a^{2}=a \sqrt{(x-c)^{2}+y^{2}} & \text { Divide by 4. } \\ {\left(c x-a^{2}\right)^{2}=a^{2}\left(\sqrt{(x-c)^{2}+y^{2}}\right)^{2}} & \text { Square both sides. } \\ c^{2} x^{2}-2 a^{2} c x+a^{4}=a^{2}\left(x^{2}-2 c x+c^{2}+y^{2}\right) & \text { Expand the squares. } \\ c^{2} x^{2}-2 a^{2} c x+a^{4}=a^{2} x^{2}-2 a^{2} c x+a^{2} c^{2}+a^{2} y^{2} & \text { Distribute } a^{2} . \\ a^{4} + c^{2} x^{2} = a^{2} x^{2} + a^{2}c^{2} + a^{2} y^{2} & \text { Combine like terms. } \\ c^{2} x^{2} - a^{2} x^{2} - a^{2} y^{2} = a^{2} c^{2} - a^{4} & \text { Rearrange terms } \\ x^{2}\left(c^{2}-a^{2}\right)-a^{2} y^{2}=a^{2}\left(c^{2}-a^{2}\right) & \text { Factor common terms. } \\ x^{2} b^{2}-a^{2} y^{2}=a^{2} b^{2} & \text { set } b^{2} = c^2 - a^2 . \\ \frac{x^{2} b^{2}}{a^{2} b^{2}}-\frac{a^{2} y^{2}}{a^{2} b^{2}}=\frac{a^{2} b^{2}}{a^{2} b^{2}} & \text { Divide both sides by } a^2 b^2 \\ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \end{array}$

STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER $(0,0)$

The standard form of the equation of a hyperbola with center $(0,0)$ and transverse axis on the $x$-axis is

$\frac{x^2}{a^2} − \frac{y^2}{b^2}=1$

where

• the length of the transverse axis is $2a$
• the coordinates of the vertices are $(±a,0)$
• the length of the conjugate axis is $2b$
• the coordinates of the co-vertices are $(0,±b)$
• the distance between the foci is $2c$, where $c^2=a^2+b^2$
• the coordinates of the foci are $(±c,0)$
• the equations of the asymptotes are $y=± \frac{b}{a}x$

See Figure 5a.

The standard form of the equation of a hyperbola with center $(0,0)$ and transverse axis on the $y$-axis is

$\frac{y^2}{a^2} − \frac{x^2}{b^2}=1$

where

• the length of the transverse axis is $2a$
• the coordinates of the vertices are $(0,±a)$
• the length of the conjugate axis is $2b$
• the coordinates of the co-vertices are $(±b,0)$
• the distance between the foci is $2c$,  where $c^2=a^2+b^2$
• the coordinates of the foci are $(0,±c)$
• the equations of the asymptotes are $y=± \frac{a}{b}x$

See Figure 5b.

Note that the vertices, co-vertices, and foci are related by the equation $c^2=a^2+b^2$. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

Figure 5 (a) Horizontal hyperbola with center $(0,0)$ (b) Vertical hyperbola with center $(0,0)$

HOW TO

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 $a^2$ 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 $\frac{x^2}{a^2}−\frac{y^2}{b^2}=1$, then the transverse axis lies on the $x$-axis. The vertices are located at  $(±a,0)$,  and the foci are located at $(±c,0)$.
   2. If the equation has the form $\frac{y^2}{a^2} − \frac{x^2}{b^2}=1$, then the transverse axis lies on the $y$-axis. The vertices are located at $(0,±a)$, and the foci are located at $(0,±c)$.
2. Solve for $a$ using the equation $a=\sqrt{a^2}$.
3. Solve for $c$ using the equation $c= \sqrt{a^2+b^2}$.

EXAMPLE 1

Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation $\frac{y^2}{49} − \frac{x^2}{32}=1$.

Solution

The equation has the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$, so the transverse axis lies on the $y$-axis. The hyerbola is centered at the origin, so the vertices serve as the $y$-intercepts of the graph. To find the vertices, set $x=0$, and solve for $y$.

$\begin{aligned} &1=\frac{y^{2}}{49}-\frac{x^{2}}{32} \\ &1=\frac{y^{2}}{49}-\frac{0^{2}}{32} \\ &1=\frac{y^{2}}{49} \\ &y^{2}=49 \\ &y=\pm \sqrt{49}=\pm 7 \end{aligned}$

The foci are located at $(0, \pm c)$. Solving for $c$,

$c=\sqrt{a^{2}+b^{2}}=\sqrt{49+32}=\sqrt{81}=9$

Therefore, the vertices are located at $(0, \pm 7)$, and the foci are located at $(0,9)$.

TRY IT #1

Identify the vertices and foci of the hyperbola with equation $\frac{x^2}{9}−\frac{y^2}{25}=1$.