Question

How do you find the standard form of the equation of the hyperbola given the properties vertices (3,2), (13,2); endpoints of the conjugate axis (8,4), (8,0)?

Answer 1

Equation of hyperbola is `(x-8)^2/25-(y-2)^2/4=1`

Explanation:

As the ordinates of vertices are equal, the equation of hyperbola is of the form `(x-h)^2/a^2-(y-k)^2/b^2=1`, where `(h,k)` is the centre of hyperbola. As vertices are `(3,2)` and `(13,2)`, centre is `((3+13)/2,(2+2)/2)` i.e. `(8,2)` and hence equation of hyperbola is

`(x-8)^2/a^2-(y-2)^2/b^2=1`

To find `a` and `b`, we will work them out using their relation with vertices and endpoints of conjugate axis. Now length of conjugate axis is `2b` and as the distance between endpoints is `4`, we have `2b=4` i.e. `b=2`.

We have the distance between vertices as `10` and as this is `2a`, we have `a=5` and hence equation of hyperbola is

`(x-8)^2/25-(y-2)^2/4=1`

graph{((x-8)^2/25-(y-2)^2/4-1)((x-13)^2+(y-2)^2-0.03)((x-3)^2+(y-2)^2-0.03)=0 [-12, 28, -7.92, 12.08]}