Question

How do you find the standard form given vertices (-6, -2) and (6,-2) and foci of (-8, -2) and (8, -2)?

Answer 1

There are two standard forms for hyperbolas:
`(x-h)^2/a^2-(y-k)^2/b^2=1" [1]"`
`(y-k)^2/a^2-(x-h)^2/b^2=1" [2]"`
Where `(h,k)` is the center point.

Explanation:

Because the vertices have the same y coordinate but different x coordinates, we know that we must use equation [1].

For equation [1] here are the general formulas for the vertices are:

`x_1 = h - a`
`x_2 = h + a`
`y_1=y_2 = k`

Substitute -6 for `x_1`, 6 for `x_2` and -2 for `y_1`:

`-6 = h - a`
`6 = h + a`
`-2 = k`

It is obvious that `h = 0`, `k = -2`, and `a = 6`.

Substitute these values into equation [1]:

`(x-0)^2/6^2-(y--2)^2/b^2=1" [3]"`

For equation [1] here are the general formulas for the foci are:

`x_1 = h - sqrt(a^2+b^2)`
`x_2 = h + sqrt(a^2+b^2)`
`y_1 = y_2 = k`

Substitute 8 for `x_2`, 0 for h and 6 for a:

`8 = 0 + sqrt(6^2+b^2)`

Solve for b:

`64 = 36+b^2`

`b^2 = 28`

`b = sqrt(28)`

Substitute this into equation [3]:

`(x-0)^2/6^2-(y--2)^2/(sqrt(28))^2=1" [4]"`

Equation [4] is the answer.