Find the standard-form equation of the hyperbola centered at the origin which satisfies the given conditions.
-Foci at $( - 20,0 ) , ( 20,0 )$; asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$
A) $\frac { y ^ { 2 } } { 256 } - \frac { x ^ { 2 } } { 144 } = 1$
B) $\frac { x ^ { 2 } } { 256 } - \frac { y ^ { 2 } } { 144 } = 1$
C) $\frac { y ^ { 2 } } { 144 } - \frac { x ^ { 2 } } { 256 } = 1$
D) $\frac { x ^ { 2 } } { 144 } - \frac { y ^ { 2 } } { 256 } = 1$

Question

Quizplus · Accepted Answer

The standard form of a hyperbola centered at the origin with foci along the x-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $2a$ is the distance between the vertices, and $2c$ is the distance between the foci, with $c^2 = a^2 + b^2$. Given foci at $(-20,0)$ and $(20,0)$, $c = 20$. The slope of the asymptotes for a hyperbola centered at the origin is $\pm \frac{b}{a}$. Given the slope of the asymptotes as $\pm \frac{3}{4}$, we have $\frac{b}{a} = \frac{3}{4}$. From the foci, we know $c = 20$, so $c^2 = 400$. Solving $c^2 = a^2 + b^2$ and $b/a = 3/4$ simultaneously, we find $a = 16$ and $b = 12$, thus $a^2 = 256$ and $b^2 = 144$. Therefore, the correct equation is $\frac{x^2}{256} - \frac{y^2}{144} = 1$.

Find the Standard-Form Equation of the Hyperbola Centered at the Origin