Graph the hyperbola. Specify the following: center, vertices, foci, lengths of transverse and conjugate axes, eccentricity, and equations of the asymptotes. $9 x ^ { 2 } - 18 x - 16 y ^ { 2 } + 96 y - 279 = 0$

Question

Quizplus · Accepted Answer

First, we need to rewrite the equation in standard form by completing the square:
$9(x-1)^2-16(y-3)^2=144$
$\frac{(x-1)^2}{16}-\frac{(y-3)^2}{9}=1$
The center is at $(1,3)$, the distance from the center to each vertex is $a=4$, so the vertices are at $(1\pm4,3)$, or $(5,3)$ and $(-3,3)$. 
The distance from the center to each focus is $c=\sqrt{a^2+b^2}$, and since $b^2=9$ we have $c=5$. So the foci are at $(6,3)$ and $(-4,3)$. 
The length of the transverse axis is $2a=8$, and the length of the conjugate axis is $2b=6$. 
The eccentricity is $e=c/a=5/4$. 
The equations of the asymptotes are $y-3=\pm\frac{3}{4}(x-1)$.
Therefore, the correct answer is (C).

Graph the Hyperbola 9x2−18x−16y2+96y−279=09 x ^ { 2 } - 18 x - 16 y ^ { 2 } + 96 y - 279 = 09x2−18x−16y2+96y−279=0

Graph the Hyperbola $9 x ^ { 2 } - 18 x - 16 y ^ { 2 } + 96 y - 279 = 0$