Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
-$$\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 4 } = 1$$
A) center at $( 0,0 )$
transverse axis is $x$-axis
vertices at $( - 11,0 )$ and $( 11,0 )$
foci at $( - \sqrt { 137 } , 0 )$ and $( \sqrt { 137 } , 0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

B) center at $( 0,0 )$
transverse axis is $\mathrm { y }$-axis
vertices at $( 0 , - 4 )$ and $( 0,4 )$
foci at $( - \sqrt { 137 } , 0 )$ and $( \sqrt { 137 } , 0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

C) center at $( 0,0 )$
transverse axis is $\mathrm { x }$-axis
vertices at $( - 4,0 )$ and $( 4,0 )$
foci at $( - 11,0 )$ and $( 11,0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

D) center at $( 0,0 )$
transverse axis is $x$-axis
vertices at $( - 4,0 )$ and $( 4,0 )$
foci at $( - \sqrt { 137 } , 0 )$ and $( \sqrt { 137 } , 0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

Question

Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
-$$\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 4 } = 1$$
A) center at $( 0,0 )$
transverse axis is $x$-axis
vertices at $( - 11,0 )$ and $( 11,0 )$
foci at $( - \sqrt { 137 } , 0 )$ and $( \sqrt { 137 } , 0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

B) center at $( 0,0 )$
transverse axis is $\mathrm { y }$-axis
vertices at $( 0 , - 4 )$ and $( 0,4 )$
foci at $( - \sqrt { 137 } , 0 )$ and $( \sqrt { 137 } , 0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

C) center at $( 0,0 )$
transverse axis is $\mathrm { x }$-axis
vertices at $( - 4,0 )$ and $( 4,0 )$
foci at $( - 11,0 )$ and $( 11,0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

D) center at $( 0,0 )$
transverse axis is $x$-axis
vertices at $( - 4,0 )$ and $( 4,0 )$
foci at $( - \sqrt { 137 } , 0 )$ and $( \sqrt { 137 } , 0 )$
asymptotes of $y = - \frac { 11 } { 4 }$ and $y = \frac { 11 } { 4 }$

Quizplus · Accepted Answer

The given equation $\frac{x^2}{36} - \frac{y^2}{4} = 1$ represents a hyperbola centered at (0,0) with a horizontal transverse axis, since the $x^2$ term is positive. The vertices are found at a distance of $\sqrt{36} = 6$ units along the x-axis from the center, so they are at $(-6,0)$ and $(6,0)$, not $(-4,0)$ and $(4,0)$ as stated in the explanation. The foci are found using the formula $c^2 = a^2 + b^2$, where $c$ is the distance from the center to a focus, $a^2 = 36$, and $b^2 = 4$, giving $c = \sqrt{40} = \sqrt{4*10} = 2\sqrt{10}$, not $\sqrt{137}$. The asymptotes for a hyperbola centered at the origin with a horizontal transverse axis are given by $y = \pm \frac{b}{a}x$, where $a = 6$ and $b = 2$, leading to $y = \pm \frac{1}{3}x$, not $y = \pm \frac{11}{4}$. Therefore, the correct details are mistakenly combined in the options, and the explanation provided here corrects those mistakes.

Find the Center, Transverse Axis, Vertices, Foci, and Asymptotes of the Hyperbola