Answer
See below
Work Step by Step
Given: $12y^2-20x^2=240$
$$\frac{y^2}{20}-\frac{x^2}{12}=1$$
The denominator of $x^2$ is greater than $y^2$, so the transverse axis is vertical.
Identify the vertices, foci, and asymptotes. Note that $a=2\sqrt 5$ and
$b=2\sqrt 3$. The $x^2-term$ is negative, so the transverse axis is vertical and the vertices are at $(0,2\pm 5)$. Find the foci:
$c^2=a^2+b^2=(2\sqrt 5)^2+(2\sqrt 3)^2=32\\
\rightarrow c=4\sqrt 2$
The foci are at $(0,\pm 4\sqrt 2)$
The asymptotes are $y=\pm \sqrt \frac{5}{3}x$
Draw the hyperbola.
