Question

How do you write the equation for a hyperbola given vertices (-5,0) and (5,0), conjugate axis of lengths 12 units?

Answer 1

Please see the explanation for steps leading to the equation:

`(x - 0)^2/5^2 - (y - 0)^2/6^2 = 1`

Explanation:

Hyperbola reference

The two vertices have the same y coordinate, therefore, the hyperbola is the horizontal transverse axis type. The standard equation for the horizontal transverse axis type is:

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

We know that the general form for the vertices of this type are:

`(h + a, k)` and `(h -a, k)`

Using the given vertices `(-5, 0)` and `(5,0)`, we can write the following equations:

`k = 0`
`h + a = 5`
`h - a = -5`

Solve for h and a:

`a = 5`
`h = 0`

From the reference, the length of the conjugate axis is equal to 2b:

`2b = 12`

`b = 6`

We know the values of h, k, a, and b, substitute them into the general form:

`(x - 0)^2/5^2 - (y - 0)^2/6^2 = 1`