A parabola is drawn with focus at one of the focii 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 axis of the Ellipse. If the latus rectum of the ellipse and that of the parabola are same, then the eccentricity of the Ellipse is

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$2 - \frac{3}{\sqrt{2}}$

$1 - \frac{1}{\sqrt{2}}$

$\sqrt{2} - 1$

$2 \sqrt{2} - 2$

answer is C.

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Detailed Solution

The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
The equation of the parabola with focus $S (a e, 0)$ and directrix $x + a e = 0$ is $y^{2} = 4 a e x$
Now, the length of latus rectum of the ellipse is $2 b^{2} / a$ and that of the parabola is $4 a e .$
For the two latus recta to be equal, we get
or $\frac{2 b^{2}}{a} = 4 a e$ or $\frac{2 a^{2} (1 - e^{2})}{a} = 4 a e$ or $1 - e^{2} = 2 e$
Therefore, $e = - \frac{2 \pm \sqrt{8}}{2} = - 1 \pm \sqrt{2}$
Hence, $e = \sqrt{2} - 1$