## Algebra and Trigonometry 10th Edition

$\frac{x^2}{4}+\frac{4y^2}{9}=1$
Major axis is horizontal. Vertices: $(2,0)=(a,0)$ $(-2,0)=(-a,0)$ $(0,\frac{3}{2})=(0,b)$ $(0,-\frac{3}{2})=(0,-b)$ That is, $a=2$ and $b=\frac{3}{2}$ Standard form when major axis is horizontal: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $\frac{x^2}{2^2}+\frac{y^2}{(\frac{3}{2})^2}=1$ $\frac{x^2}{4}+\frac{y^2}{\frac{9}{4}}=1$ $\frac{x^2}{4}+\frac{4y^2}{9}=1$