College Algebra, Chapter 2, 2.2, Section 2.2, Problem 38

Find the $x$ and $y$ intercept of $\displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1$


$
\begin{equation}
\begin{aligned}

\frac{x^2}{9} + \frac{y^2}{4} =& 1
&& \text{Given}
\\
\\
4x^2 + 9y^2 =& 36
&& \text{Multiply both sides by LCD } 36

\end{aligned}
\end{equation}
$


To solve for $x$ intercept, we set $y = 0$


$
\begin{equation}
\begin{aligned}

4x^2 + 9(0)^2 =& 36
\\
\\
4x^2 =& 36
\\
\\
x^2 =& 9
\\
\\
x =& \pm 3

\end{aligned}
\end{equation}
$


The $x$ intercepts are at $(3, 0)$ and $(-3, 0)$

To solve for $y$ intercept, we set $x =0$


$
\begin{equation}
\begin{aligned}

4(0)^2 + 9(y^2) =& 36
\\
\\
9(y^2) =& 36
\\
\\
y^2 =& 4
\\
\\
y =& \pm 2

\end{aligned}
\end{equation}
$


The $y$ intercepts are at $(0,2)$ and $(0, -2)$