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

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

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Answer 1 · 2016-11-11T22:24:25

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:

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

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

1 Answer

Explanation:

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