Question

Let $x+y=0$ and $2x-y+3=0$ be the major and minor axis of an ellipse respectively. If the foot of perpendicular drawn from vertex of the parabola $x^2-4x+4y+16=0$ to these lines is focus and one endpoint of minor axis respectively, then the eccentricity of the ellipse is

• A. $\frac{7}{\sqrt{59}}$
• B. $\frac{7}{8}$
• C. $\frac{5}{\sqrt{59}}$
• D. $\frac{4}{\sqrt{53}}$

Answer 1

The correct option is A $\frac{7}{\sqrt{59}}$

Centre of the ellipse is point of intersection of $x+y=0$ and $2x-y+3=0$
$C=(-1,1)$


Let $P$ be the vertex of parabola
$x^2-4x+4y+16=0\Rightarrow (x-2{)}^2=-4(y+3)\Rightarrow P=(2,-3)$
Foot of perpendicular from $P$ to $x+y=0$ is
$\frac{x-2}{1}=\frac{y+3}{1}=\frac{-(-1)}{2}\therefore S=(x,y)=(\frac{5}{2},-\frac{5}{2})$

Distance from centre $C$ to focus $S$ is
$CS=ae\Rightarrow ae=\frac{7}{\sqrt{2}}\cdots \left(1\right)$

Foot of perpendicular from $P$ to $2x-y+3=0$ is
$\frac{x-2}{2}=\frac{y+3}{-1}=\frac{-\left(10\right)}{5}\therefore B=(x,y)=(-2,-1)$

Distance from centre $C$ to one endpoint of minor axis $B$ is
$\Rightarrow CB=b\Rightarrow b=\sqrt{5}\cdots \left(2\right)$

We know that
$e^2=1-\frac{b^2}{a^2}$
Using equations $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow e^2=1-\frac{5\times 2e^2}{49}\Rightarrow \frac{59}{49}{e}^2=1\therefore e=\frac{7}{\sqrt{59}}$