College Algebra, Chapter 8, 8.2, Section 8.2, Problem 46

Determine the parts of intersection of the pair of ellipses.

$
\left\{
\begin{equation}
\begin{aligned}
\frac{x^2}{16} + \frac{y^2}{9} &= 1\\
\\
\frac{x^2}{9} + \frac{y^2}{16} &= 1
\end{aligned}
\end{equation}
\right.
$

Sketch the graphs on the same axes and label the points of intersection.
By using elimination method,

$
\left\{
\begin{equation}
\begin{aligned}
\frac{x^2}{144} + \frac{y^2}{81} &= \frac{1}{9} && \frac{1}{9} \times \text{Equation 1}\\
\\
-\frac{x^2}{144} - \frac{y^2}{256} &= -\frac{1}{16} && -\frac{1}{16} \times \text{Equation 2}
\end{aligned}
\end{equation}
\right.
$


$
\qquad
\begin{equation}
\begin{aligned}
\hline\\
\frac{y^2}{81} - \frac{y^2}{256} &= \frac{1}{9} - \frac{1}{16} && \text{Add}\\
\\
\frac{175y^2}{20736} &= \frac{7}{144} && \text{Simplify}\\
\\
y &= \pm \frac{12}{5} = \pm 2.4
\end{aligned}
\end{equation}
$

Now we back substitute the value of $y$ in either of the equation. So if $\displaystyle y = \pm\frac{12}{5}$, then

$
\begin{equation}
\begin{aligned}
\frac{x^2}{16} + \frac{\left( \pm \frac{12}{5} \right)^2}{9} &= 1\\
\\
x &= \pm \frac{12}{5} = \pm 2.4
\end{aligned}
\end{equation}
$

Therefore, the parts of intersections are $(2.4, 2.4), (2.4,-2.4),(-2.4,2.4)$ and $(-2.4,-2.4)$