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

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

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

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

Quizplus · Accepted Answer

The equation of the hyperbola is in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, indicating the transverse axis is along the x-axis. The center is at (0,0) because there are no shifts in the equation. The vertices are at $\pm a = \pm 6$ units along the x-axis from the center, so at (-6,0) and (6,0), not (-4,0) and (4,0) as stated in the choices. The foci are found using $c^2 = a^2 + b^2$, giving $c = \sqrt{36 + 4} = \sqrt{40}$, not $\sqrt{137}$ or 11. The asymptotes for a hyperbola centered at the origin are given by $y = \pm \frac{b}{a}x$, which in this case are $y = \pm \frac{1}{3}x$, not $y = \pm \frac{11}{4}$. Therefore, none of the provided choices accurately describe all the elements of the given hyperbola.

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