Answer
$\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
Foci: $\qquad (\pm 5,0)$
Directrices: $\quad x=\displaystyle \pm\frac{16}{5}$
Work Step by Step
Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$
Write in standard form and read a,b.
$9x^{2}-16y^{2}=144\qquad /\div 144$
$\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1\qquad$
We have a horizontal axis, $a=4, b=3$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{16+9}=5$
Foci: $\quad (\pm c, 0)=\qquad (\pm 5,0)$
Vertices: $\quad (\pm a, 0)=\qquad (\pm 4, 0)$
Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x \Rightarrow \quad y=\displaystyle \pm\frac{3}{4}x$
Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{5}{4}$
Directrices: $\quad x=0\displaystyle \pm\frac{a}{e}=\pm\frac{16}{5}$
