How do you find the equation for a hyperbola centered at the origin with a horizontal transverse axis of lengths 8 units and a conjugate axis of lengths 6 units?

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Answer 1 · 2016-12-08T19:01:51

The equation is: $\frac{{(x - 0)}^{2}}{4^{2}} - \frac{{(y - 0)}^{2}}{3^{2}} = 1$ $(x - 0)^2/4^2 - (y - 0)^2/3^2 = 1$

Using this reference Conics: Hyperbola, the general equation for a hyperbola with a transverse horizontal axis is:

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

where $(h, k)$ $(h, k)$ is the center, $2 a$ $2a$ is the length of transverse axis, and $2 b$ $2b$ is the length of the conjugate axis.

We are given that the center is, $(0, 0)$ $(0, 0)$ ; substitute this into equation [1]:

$\frac{{(x - 0)}^{2}}{a^{2}} - \frac{{(y - 0)}^{2}}{b^{2}} = 1 [2]$ $(x - 0)^2/a^2 - (y - 0)^2/b^2 = 1" [2]"$

We are given that the length of the horizontal transverse axis is 8; this allows us to write this equation:

$2 a = 8$ $2a = 8$

$a = 4$ $a = 4$

Substitute 4 for "a" in equation [2]:

$\frac{{(x - 0)}^{2}}{4^{2}} - \frac{{(y - 0)}^{2}}{b^{2}} = 1 [3]$ $(x - 0)^2/4^2 - (y - 0)^2/b^2 = 1" [3]"$

We are given that the length of the horizontal transverse axis is 6; this allows us to write this equation:

$2 b = 6$ $2b = 6$

$b = 3$ $b = 3$

Substitute 3 for "b" in equation [3]:

$\frac{{(x - 0)}^{2}}{4^{2}} - \frac{{(y - 0)}^{2}}{3^{2}} = 1 [4]$ $(x - 0)^2/4^2 - (y - 0)^2/3^2 = 1" [4]"$