## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

${{x}^{2}}+9{{y}^{2}}=9$ …… (1) Multiply $\frac{1}{9}$ on both the sides of the equation ${{x}^{2}}+9{{y}^{2}}=9$. Identifying $a\text{ and }b$ in the equation. \begin{align} & \frac{{{x}^{2}}}{9}+\frac{9}{9}{{y}^{2}}=\frac{9}{9} \\ & \frac{{{x}^{2}}}{9}+{{y}^{2}}=1 \\ & \frac{{{x}^{2}}}{{{\left( 3 \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( 1 \right)}^{2}}}=1 \end{align} Since ${{a}^{2}}>{{b}^{2}}$, the ellipse will be horizontal. Since $a=3\text{ and }b=1$, the x-intercepts are $\left( -3,0 \right)\text{ and }\left( 3,0 \right)$ and the y-intercepts are $\left( 0,-1 \right)\text{ and }\left( 0,1 \right)$. Plot this point and connect them with the oval shaped curve.