How do you find the equation for a hyperbola centered at the origin with a vertical transverse axis of lengths 12 units and a conjugate axis of length 4 units?

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How do you find the equation for a hyperbola centered at the origin with a vertical transverse axis of lengths 12 units and a conjugate axis of length 4 units?

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Answer 1 · 2017-12-15T21:22:28

The general form for a the equation of a hyperbola with a vertical transverse axis is:

$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$ $(y-k)^2/a^2-(x-h)^2/b^2=1$

Explanation:

The center of the general form is $(h, k)$ $(h,k)$ . We are told that the center is the origin: $(0, 0)$ $(0,0)$ , therefore, we can remove h and k from the equation:

$\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ $y^2/a^2-x^2/b^2=1$

The length of the transverse axis is $2 a$ $2a$

$2 a = 12$ $2a = 12$

$a = 6$ $a = 6$

$\frac{y^{2}}{6^{2}} - \frac{x^{2}}{b^{2}} = 1$ $y^2/6^2-x^2/b^2=1$

The length of the conjugate axis is $2 b$ $2b$ :

$2 b = 4$ $2b = 4$

$b = 2$ $b = 2$

$\frac{y^{2}}{6^{2}} - \frac{x^{2}}{2^{2}} = 1$ $y^2/6^2-x^2/2^2=1$

Done.

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How do you find the equation for a hyperbola centered at the origin with a vertical transverse axis of lengths 12 units and a conjugate axis of length 4 units?

1 Answer

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