Question

Find the equation of the hyperbola with vertices at (0, ± 6) and e = `5/3`. Find its foci.

Answer 1

Since the vertices are on the y-axes (with origin at the mid-point)

The equation is of the form `y^2/a^2 - x^2/b^2` = 1.

As vertices are (0, ± 6)

a = 6

b² = a²(e² – 1)

= `36  25/9 - 1`

= 64

So the required equation of the hyperbola is `y^2/36 - x^2/64` = 1

And the foci are (0, ± ae) = (0, ± 10).