Algebra and Trigonometry 10th Edition

$\frac{x^2}{4}-\frac{5y^2}{12}=1$
Transverse axis is horizontal. $(±a,0)=(±2,0)$ $a=2$ Use the point $(3,\sqrt 3)$: Standard form when transverse axis is horizontal: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $\frac{3^2}{2^2}-\frac{(\sqrt 3)^2}{b^2}=1$ $\frac{9}{4}-\frac{3}{b^2}=1$ $\frac{9}{4}-1=\frac{3}{b^2}$ $\frac{5}{4}=\frac{3}{b^2}$ $b^2=\frac{12}{5}$ $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $\frac{x^2}{4}-\frac{y^2}{\frac{12}{5}}=1$ $\frac{x^2}{4}-\frac{5y^2}{12}=1$