Answer
$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
Work Step by Step
$a^2-b^2=4^2$
$a^2-b^2=16$
$\frac{(-4)^{2}}{a^2}+\frac{(1.8)^{2}}{b^2}=1$
$16b^2+3.24a^2=a^2b^2$
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,
$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
