Question

One focus and the corresponding directrix of an ellipse are $(1,2)$ and $x-y=5$. Its eccentricity is $\frac{1}{2}$. Then its centre is:

• A. $(3,0)$
• B. $(0,3)$
• C. $(3,-3)$
• D. $(0,-3)$

Answer 1

The correct option is B $(0,3)$

Given one focus is $(1,2)$ and directrix is $x-y=5$ and eccentricity $e=\frac{1}{2}$. If point $P$ is on the ellipse then $SP=ePM$

$SP$=distance between focus and $P$

$PM$=Perpendicular distance on directrix.

$A,A^{\prime }$ are points on the ellipse. $SA=eAM$ and $S{A}^{\prime }=eAM$.

So, $A$ divides $SM$ in $e:1$ internally.

$A^{\prime }$ divides $SM$ in $e:1$ externally.

Major axis is $\frac{y-2}{x-1}=-1\Rightarrow y=-x+3$

M is the intersection of major axis and directrix which is $(4,-1)$.

$A=(\frac{\frac{1}{2}\times 4+1\times 1}{\frac{3}{2}},\frac{\frac{1}{2}\times -1+1\times 2}{\frac{3}{2}})=(2,1)$

$A^{\prime }=(\frac{\frac{1}{2}\times 4-1\times 1}{\frac{-1}{2}},\frac{\frac{1}{2}\times -1-1\times 2}{\frac{-1}{2}})=(-2,5)$

midpoint of $A{A}^{\prime }$ is the centre of the ellipse, centre=$(\frac{2-2}{2},\frac{1+5}{2})=(0,3)$