Question

Find the equation of the ellipse whose focus is $(1,-1)$ eccentricity is $\frac{2}{3}$ and directrix is $x+y+2=0$.

Answer 1

Let $F$ be the given focus of ellipse and $P(h,k)$ be a point on the ellipse.

$\therefore$ distance $FP=\sqrt{(h-1{)}^2+(k+1{)}^2}$.

Also, shortest distance of $P$ from directrix = $PD=\left|\frac{h+k+2}{\sqrt{1^2+1^2}}\right|=\left|\frac{h+k+2}{2}\right|$

Now, from definition of eccentricity,

$e=\frac{FP}{DP}$

$\therefore \frac{2}{3}=\frac{\sqrt{(h-1{)}^2+(k+1{)}^2}}{\left|\frac{h+k+2}{2}\right|}$

$\therefore \frac{2}{3}\left|\frac{h+k+2}{2}\right|=\sqrt{(h-1{)}^2+(k+1{)}^2}$

Squaring both sides,

$\frac{4}{9}{\left(\frac{h+k+2}{2}\right)}^2=(h-1{)}^2+(k+1{)}^2$

$\therefore (h+k+2{)}^2=9(h-1{)}^2+9(k+1{)}^2$

$\therefore h^2+k^2+4+2hk+4h+4k=9h^2-18h+9+9k^2+18k+9$

$\therefore 8h^2+8k^2-2hk-22h+14k+14=0$

Hence, the required equation of ellipse is $8x^2+8y^2-2xy-22x+14y+14=0$.