Question

The focus of an ellipse is $M (-1,-1)$ and corresponding directrix $x - y +3=0$ and eccentricity is $\frac{1}{2}$. If the centre of the ellipse be $C(p, q)$, then find the value of $(2 p-4 q)$.

Answer 1

We have focus $(-1,-1)$ and corresponding directrix $x-y+3=0$ ....(1)
$e =\frac{1}{2} \text { (given) }$
Now, equation of axis of ellipse is
$x+y+k=0$
As (1) passes through $(-1,-1)$, so $k =2$
$\therefore $ Equation of axis is
$x + y +2=0$ ....(2)
$\therefore $ On solving (1) and (3), we get foot of directrix $\left(\frac{-5}{2}, \frac{1}{2}\right)$.
Now, distance between focus and corresponding foot of directrix
$=\left(\frac{ a }{ e }- ae \right)=\sqrt{\left(\frac{-5}{2}+1\right)^2+\left(\frac{1}{2}+1\right)^2} \left(\text { As e }=\frac{1}{2}\right)$
$\Rightarrow 2 a -\frac{ a }{2}=\frac{3}{\sqrt{2}} \Rightarrow a =\sqrt{2}$
Also equation of minor axis is
$x-y+\lambda=0$
As distance of minor axis from focus is ae, so
$\frac{|-1+1+\lambda|}{\sqrt{2}}=\sqrt{2}\left(\frac{1}{2}\right) \Rightarrow|\lambda|=1 \Rightarrow \lambda= \pm 1$
As focus lies between minor axis and corresponding diretrix, so equation of minor axis is
$x - y -1=0$
$\therefore $ On solving (3) and (4), we get
$x=\frac{-1}{2}, y=\frac{-3}{2} $
$\therefore \text { Centre } \equiv\left(\frac{-1}{2}, \frac{-3}{2}\right) \equiv(p, q)$
Hence $(2 p-4 q)=-1+6=5$