How do you write a standard form equation for the hyperbola with vertices at (-2, 0) and (2, 0) and foci at (-6, 0) and (6, 0)?

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Answer 1 · 2017-05-26T20:29:39

Because the vertices are horizontal, we know that the standard form is,

$(x-h)^2/a^2-(y-k)^2/b^2=1" [1]"$

, the vertices are $(h+-a,k)$

and the foci are $(h+-sqrt(a^2+b^2),k)$

Using the form of the vertices and the given vertices we can write the following equations:

$-2 = h-a$
$2 = h+a$
$k = 0$

Solving the first two equations we have:

$h = 0$
$a = 2$
$k =0$

Using the form of the foci and one of the given foci we can write:

$6 = h + sqrt(a^2+b^2)$

Substitute in the known values for h and a:

$6 = 0 + sqrt(2^2+b^2)$

Solve for b:

$36=4+b^2$

$b = sqrt32$

Substituting the know values into equation [1]:

$(x-0)^2/2^2-(y-0)^2/(sqrt32)^2=1" [2]"$

Equation [2] is the answer.