Answer
See below
Work Step by Step
Given: $81x^2-16y^2=1296$
Rewrite the equation in standard form
$$\frac{x^2}{16}-\frac{y^2}{81}=1$$
The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is vertical.
Identify the vertices, foci, and asymptotes. Note that $a=4$ and
$b=9$. The $x^2-term$ is positive, so the transverse axis is vertical and the vertices are at $(\pm 4,0)$. Find the foci:
$c^2=a^2+b^2=4^2+9^2=97\\
\rightarrow c=\sqrt 97$
The foci are at $(\pm \sqrt 97,0)$
The asymptotes are $y=\pm \frac{9}{4}x$
Draw the hyperbola.
