Question

Find the equation of the hyperbola whose vertices are (± 6, 0) and one of the directrices is x = 4.

Answer 1

As the vertices are on the x-axis and their middle point is the origin

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

Here b² = a²(e² – 1)

Vertices are (± a, 0)

And Directrices are given by x = `+-  a/e`.

Thus `a= 6, a/e` = 4

And so e = `3/2`

Which gives b² = `36  9/4 - 1` = 45

Consequently, the required equation of the hyperbola is `x^2/36 - y^2/45` = 1