Answer
$\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{64}=1$
Foci: $\qquad (\pm 10,0)$
Directrices: $\quad x=\displaystyle \pm\frac{18}{5}$
Work Step by Step
Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$
Write in standard form and read a,b.
$64x^{2}-36y^{2}=2304\qquad /\div 2304$
$\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{64}=1\qquad $
We have a horizontal axis, $a=6, b=8$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{36+64}=10$
Foci: $\quad (\pm c, 0)=\qquad (\pm 10,0)$
Vertices: $\quad (\pm a, 0)=\qquad (\pm\sqrt{2}, 0)$
Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x \Rightarrow \quad y=\displaystyle \pm\frac{4}{3}x$
Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$
Directrices: $\quad x=0\displaystyle \pm\frac{a}{e}=\pm\frac{6}{\frac{5}{3}}=\pm\frac{18}{5}$
