Find the vertices and locate the foci for the hyperbola whose equation is given.
-$\frac { \mathrm { x } ^ { 2 } } { 144 } - \frac { \mathrm { y } ^ { 2 } } { 4 } = 1$
A) vertices: $( - 12,0 ) , ( 12,0 )$
foci: $( - 2 \sqrt { 37 } , 0 ) , ( 2 \sqrt { 37 } , 0 )$
B) vertices: $( - 2,0 ) , ( 2,0 )$
foci: $( - 2 \sqrt { 37 } , 0 ) , ( 2 \sqrt { 37 } , 0 )$
C) vertices: $( 0 , - 12 ) , ( 0,12 )$
foci: $( - 2 \sqrt { 37 } , 0 ) , ( 2 \sqrt { 37 } , 0 )$
D) vertices: $( - 12,0 ) , ( 12,0 )$
foci: $( - 2,0 ) , ( 2,0 )$

Question

Quizplus · Accepted Answer

We can rewrite the given equation as $\frac{(x-0)^2}{12^2}-\frac{(y-0)^2}{2^2}=1$, which is in the form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ where $(h,k)=(0,0)$. Thus, we know that the center of the hyperbola is at $(0,0)$. 
The vertices are located on the transverse axis, which is the line passing through the center and perpendicular to the conjugate axis. In this case, the transverse axis is the $x$-axis, so the vertices are located at $(\pm a,0)=(\pm 12,0)$. Thus, the answer choices (A) and (D) could both be correct. 
To find the foci, we use the formula $c=\sqrt{a^2+b^2}$, where $c$ is the distance from the center to each focus. Plugging in $a=12$ and $b=2$, we get $c=\sqrt{12^2+2^2}=2\sqrt{37}$. Therefore, the foci are located at $(\pm c,0)=(\pm 2\sqrt{37},0)$. Choice (A) is the only one that includes both the correct vertices and foci.

Find the Vertices and Locate the Foci for the Hyperbola x2144−y24=1\frac { \mathrm { x } ^ { 2 } } { 144 } - \frac { \mathrm { y } ^ { 2 } } { 4 } = 1144x2​−4y2​=1

Find the Vertices and Locate the Foci for the Hyperbola $\frac { \mathrm { x } ^ { 2 } } { 144 } - \frac { \mathrm { y } ^ { 2 } } { 4 } = 1$