## Calculus 8th Edition

$\dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1$
We have $a^2-b^2=4^2=16$ or $a^2=16+b^2$ ... (1) Thus, $\dfrac{(-4)^{2}}{a^2}+\dfrac{(1.8)^{2}}{b^2}=1$ or, $16b^2+(3.24)a^2=a^2b^2$ or, $b^4-3.24b^2-51.84=0$ Thus, $b^2=9$ From equation (1), we have $a^2=16+b^2$ or, $a^2=16+9=25$ Hence, $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1$