Answer
The graph is shown below
Work Step by Step
$16{{x}^{2}}+25{{y}^{2}}=1$
Now write $16=\frac{1}{\frac{1}{16}}$ and $25=\frac{1}{\frac{1}{25}}$, so rewrite the equation.
$\frac{{{x}^{2}}}{\frac{1}{16}}+\frac{{{y}^{2}}}{\frac{1}{25}}=1$
Standard equation of an Ellipse is:
$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$
Identify $a\text{ and }b$ by comparing the equation with the standard equation,
$\begin{align}
& \frac{{{x}^{2}}}{\frac{1}{16}}+\frac{{{y}^{2}}}{\frac{1}{25}}=1 \\
& \frac{{{x}^{2}}}{{{\left( \frac{1}{4} \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( \frac{1}{5} \right)}^{2}}}=1
\end{align}$
Since $a=\frac{1}{4}\text{ and }b=\frac{1}{5}$, the x-intercepts are $\left( -\frac{1}{4},0 \right)\text{ and }\left( \frac{1}{4},0 \right)$ and the y-intercepts are $\left( 0,-\frac{1}{5} \right)\text{ and }\left( 0,\frac{1}{5} \right)$.
Since ${{a}^{2}}>{{b}^{2}}$, the ellipse will be horizontal.
Plot this point and connect them with the oval shaped curve.
