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

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Answer 1 · 2017-04-08T19:30:28

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.

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.