What is the equation of the hyperbola with a center at (0, 0), a vertex at (0, 60), and a focus at (0, -65)?

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What is the equation of the hyperbola with a center at (0, 0), a vertex at (0, 60), and a focus at (0, -65)?

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Answer 1 · 2018-01-14T00:59:48

$\frac{x^{2}}{25^{2}} - \frac{y^{2}}{60^{2}} = - 1$ $x^2/25^2 - y^2/60^2 =-1$

Explanation:

This hyperbola has a center at $(0, 0)$ $(0,0)$ , and its vertexes and foci are on the $y$ $y$ -axis.
Therefore, the equation of the hyperbola must be in the form
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ $x^2/a^2 - y^2/b^2 =-1$ . $(a > 0, b > 0)$ $(a>0, b>0)$

In the equation, vertexes are $(0, \pm b)$ $(0, +-b)$ . So, $b = 60$ $b=60$ .
Foci are $(0, \pm \sqrt{a^{2} + b^{2}})$ $(0,+-sqrt(a^2+b^2))$ , and you need to solve
$\sqrt{a^{2} + 60^{2}} = 65$ $sqrt(a^2+60^2)=65$
$a^{2} = 65^{2} - 60^{2} = 625 = 25^{2}$ $a^2=65^2-60^2=625=25^2$ .

The result is $\frac{x^{2}}{25^{2}} - \frac{y^{2}}{60^{2}} = - 1$ $x^2/25^2 - y^2/60^2 =-1$ .

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What is the equation of the hyperbola with a center at (0, 0), a vertex at (0, 60), and a focus at (0, -65)?

1 Answer

Explanation:

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