Answer
See below
Work Step by Step
Given: $4x^2-y^2=256$
Rewrite the equation in standard form
$$\frac{x^2}{64}-\frac{y^2}{256}=1$$
The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is vertical.
Identify the vertices, foci, and asymptotes. Note that $a=8$ and
$b=16$. The $x^2-term$ is positive, so the transverse axis is vertical and the vertices are at $(0,\pm 7)$. Find the foci:
$c^2=a^2+b^2=16^2+8^2=320\\
\rightarrow c=8\sqrt 5$
The foci are at $(\pm 8\sqrt 5,0)$
The asymptotes are $y=\pm 2x$
Draw the hyperbola.
