Question

A parabola is drawn with focus at one of the foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (where $a>b$) and directrix passing through the other focus and perpendicular to the major axes of the ellipse. If latus rectum of the ellipse and the parabola are same, then the eccentricity of the ellipse is

• A. $1-\frac{1}{\sqrt{2}}$
• B. $2\sqrt{2}-2$
• C. $\sqrt{2}-1$
• D. None of these

Answer 1

The correct option is C $\sqrt{2}-1$

Equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Equation of the parabola with focus $S(ae,0)$ and directrix $x+ae=0$ is $y^2=4aex$

Now length of latus rectum of the ellipse is $\frac{2b^2}{a}$ and that of the parabola is $4ae$.

For the two latus rectum to be equal, we get

$\frac{2b^2}{a}=4ae$

$\Rightarrow \frac{2a^2(1-e^2)}{a}=4ae$

$\Rightarrow 1-e^2=2e$

$\Rightarrow e^2+2e-1=0$

Therefore, $e=-\frac{2\pm \sqrt{8}}{2}=-1\pm \sqrt{2}$

Hence, $e=\sqrt{2}-1$