Answer
$\displaystyle \frac{x^{2}}{\frac{64}{3}}+\frac{y^{2}}{\frac{16}{3}}=1$
Work Step by Step
See fig. 11.47 $\quad $
Foci = $(\pm c,0)\quad \Rightarrow c=4.$
$c=ae=4\displaystyle \quad \Rightarrow a=\frac{4}{e}$
If directrix 2 is $x=\displaystyle \frac{16}{3}$, then $\displaystyle \frac{a}{e}=\frac{16}{3}$
Substituting $a=\displaystyle \frac{4}{e}$, we have
$\displaystyle \frac{4}{e^{2}}=\frac{16}{3}$
$12=16e^{2}$
$e^{2}=\displaystyle \frac{12}{16}$
$e^{2}=\displaystyle \frac{3}{4}$
$e=\displaystyle \frac{\sqrt{3}}{2}$
If $P$ is a point on the ellipse, then $|PF_{2}|=a|PD_{2}|$ (using fig. 11.47 )
$\displaystyle \sqrt{(x-4)^{2}+(y-0)^{2}}=\frac{\sqrt{3}}{2}|x-\frac{16}{3}|$
$(x-4)^{2}+y^{2}=\displaystyle \frac{3}{4}(x-\frac{16}{3})^{2}$
$x^{2}-8x+16+y^{2}=\displaystyle \frac{3}{4}(x^{2}-\frac{32}{3}x+\frac{256}{9})$
$x^{2}-8x+16+y^{2}=\displaystyle \frac{3}{4}x^{2}-8x+\frac{64}{3}$
$\displaystyle \frac{1}{4}x^{2}+y^{2}=\frac{64-48}{3}$
$\displaystyle \frac{1}{4}x^{2}+y^{2}=\frac{16}{3}\qquad/\times\frac{3}{16}$
$\displaystyle \frac{x^{2}}{\frac{64}{3}}+\frac{y^{2}}{\frac{16}{3}}=1$
