Question

Find the equation of the hyperbola that has its foci at (0,-4) ans (10,-4) and whose conjugate axis is 6 units long

Answer 1

We are given the following information

Foci at (0, -4) and (10, -4)

The conjugate axis is 6 units

Recall that the standard form of the equation of hyperbola centered at (h, k) is given by

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

Where the center is (h, k)

The center (h, k) is the midpoint of the transverse axis given by

$\begin{gathered} (h,k)=(\frac{0+10}{2},\frac{-4+(-4)}{2}) \\ (h,k)=(\frac{0+10}{2},\frac{-4-4}{2}) \\ (h,k)=(\frac{10}{2},\frac{-8}{2}) \\ (h,k)=(5,-4) \end{gathered}$

Let us find the values of a and b

The semi transverse axis (a) is given by

$a=\frac{10-0}{2}=\frac{10}{2}=5$

The semi conjugate axis (b) is given by

$b=\frac{6}{2}=3$

So, the equation of the hyperbola is

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \frac{(x-5)^2}{5^2}-\frac{(y-(-4))^2}{3^2}=1 \\ \frac{(x-5)^2}{25^{}}-\frac{(y+4)^2}{9^{}}=1 \end{gathered}$

Therefore, the equation of the hyperbola is

$\frac{(x-5)^2}{25^{}}-\frac{(y+4)^2}{9^{}}=1$