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

`((x-h)^2)/(a^2)-((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)=((0+10)/(2),(-4+(-4))/(2)) \\ (h,k)=((0+10)/(2),(-4-4)/(2)) \\ (h,k)=((10)/(2),(-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=(10-0)/(2)=(10)/(2)=5`

The semi conjugate axis (b) is given by

`b=(6)/(2)=3`

So, the equation of the hyperbola is

`\begin{gathered} ((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1 \\ ((x-5)^2)/(5^2)-((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`