Answer
See below
Work Step by Step
Given: $49x^2-4y^2=196$
Rewrite the equation in standard form
$$\frac{x^2}{4}-\frac{y^2}{49}=1$$
The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is vertical.
Identify the vertices, foci, and asymptotes. Note that $a=2$ and
$b=7$. The $x^2-term$ is positive, so the transverse axis is vertical and the vertices are at $(\pm 7,0)$. Find the foci:
$c^2=a^2+b^2=2^2+7^2=53\\
\rightarrow c=\sqrt 53$
The foci are at $(\pm \sqrt 53,0)$
The asymptotes are $y=\pm \frac{7}{2}x$
Draw the hyperbola.
