Question

Find an equation of the hyperbola having foci at (-3, 3) and (-3,9) and vertices at (-3, 4) and (-3,8).

Answer 1

Given:

Foci of the hyperbola = (-3,3), (-3, 9)

Vertices of the hyperbola = (-3, 4), (-3, 8)

Required: Equation of the hyperbola

Explanation:

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

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

First, identify the center (h, k). The center is halfway between the vertices (-3, 4) and (-3, 8). Apply the midpoint formula.

$\begin{gathered} (h,k)=(\frac{-3+-3}{2},\frac{4+8}{2}) \\ =(-3,6) \end{gathered}$

Next, find the square of a.

The length of the transverse axis is, 2a, bounded by the vertices. So, to find the square of a, determine the distance between the y-coordinates of the vertices.

$\begin{gathered} 2a=|4-8| \\ 2a=4 \\ a=\frac{4}{2}=2 \\ a^2=2^2=4 \end{gathered}$

Find the square of c. The coordinates of the foci are (h, k+c) and (h, k-c). So (h, k- c) = (-3, 3) and (h, k+c) = (-3, 9). Use the y-coordinate from either of these points to solve for c. Using the point (-3, 3) and substituting k = 6.

$\begin{gathered} k+c=9 \\ 6+c=9 \\ c=9-6 \\ =3 \\ c^2=3^2=9 \end{gathered}$

Solve for square of b using the equation

$b^2=c^2-a^2$

Substitute the obtained values.

$\begin{gathered} b^2=9-4 \\ =5 \end{gathered}$

Finally, substitute the obtained values into the standard form of the equation

$\begin{gathered} \frac{(y-6)^2}{4}-\frac{(x-(-3))^2}{9}=1 \\ \frac{(y-6)^{2}}{4}-\frac{(x+3)^2}{9}=1 \end{gathered}$

Final Answer: The equation of the required hyperbola is

$\frac{(y-6)^2}{4}-\frac{(x+3)^2}{9}=1$