## Calculus: Early Transcendentals 8th Edition

$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
$a^2-b^2=4^2$ $a^2-b^2=16$ $\frac{(-4)^{2}}{a^2}+\frac{(1.8)^{2}}{b^2}=1$ $16b^2+3.24a^2=a^2b^2$ use $a^2-b^2=16$ Thus, $b^4-3.24b^2-51.84=0$ $b=3$ Now, $a^2=16+b^2$ $a=5$ Thus, $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$