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 $(2p-4q)$.

Answer 1

Answer: 0005

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
$=\left(\frac{a}{e}-ae\right)=\sqrt{{\left(\frac{-5}{2}+1\right)}^2+{\left(\frac{1}{2}+1\right)}^2}\left(\text{ As e }=\frac{1}{2}\right)$
$\Rightarrow 2a-\frac{a}{2}=\frac{3}{\sqrt{2}}\Rightarrow a=\sqrt{2}$
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 $(2p-4q)=-1+6=5$