Find the center, foci, and asymptotes of the hyperbola.
-$$x ^ { 2 } - y ^ { 2 } = 128$$
A) C: $( 0,0 ) ; F : ( 16,0 ) , ( - 16,0 ) ; A : y = x , y = - x$
B) C: $( 0,0 ) ; \mathrm { F } : ( 8,0 ) , ( - 8,0 ) ; \mathrm { A } : \mathrm { y } = \mathrm { x } , \mathrm { y } = - \mathrm { x }$
C) C: $( 2,2 ) ; F : ( 16,0 ) , ( - 16,0 ) ; A : y = 2 x , y = - 2 x$
D) $\mathrm { C } : ( 0,0 ) ; \mathrm { F } : ( 16,0 ) , ( - 16,0 ) ; \mathrm { A } : \mathrm { y } = 2 \mathrm { x } , \mathrm { y } = - 2 \mathrm { x }$

Question

Quizplus · Accepted Answer

The given equation $x^2 - y^2 = 128$ is in the standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a^2 = 128$, so $a = \sqrt{128} = 8\sqrt{2}$. The center is at the origin $(0,0)$ because there are no terms to shift the hyperbola. The foci are found using $c^2 = a^2 + b^2$, but since $b^2 = 0$ in this case (indicating a horizontal transverse axis), $c = a = 8\sqrt{2}$, so the foci are $(\pm 8\sqrt{2}, 0)$, which simplifies to $(\pm 16, 0)$. The asymptotes of a hyperbola of this form are $y = \pm \frac{b}{a}x$, but since $a = b$ here, the asymptotes are $y = \pm x$.

Find the Center, Foci, and Asymptotes of the Hyperbola x2−y2=128x ^ { 2 } - y ^ { 2 } = 128x2−y2=128

Find the Center, Foci, and Asymptotes of the Hyperbola $x ^ { 2 } - y ^ { 2 } = 128$