Answer
See below
Work Step by Step
Given $$\frac{x^2}{16}+\frac{y^2}{49}=1$$
The equation in standard form
The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is vertical.
Identify the vertices, foci, and asymptotes. Note that $a=7$ and
$b=4$. The $x^2-term$ is negative, so the transverse axis is vertical and the vertices are at $(0,\pm 10)$. Find the foci:
$c^2=a^2-b^2=7^2-4^2=33\\
\rightarrow c=\sqrt 33$
The foci are at $(0,\pm \sqrt 33)$
