Question

A parabola is drawn whose focus is one of the foci of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1(where a > b) and whose directrix passes through the other focus and perpendicular to the major axes of the ellipse. Then the eccentricity of the ellipse for which the latus-rectum of the ellipse and the parabola are same, is –

• A. $\sqrt{2}$ – 1
• B. 2$\sqrt{2}$+ 1
• C. $\sqrt{2}$ + 1
• D. 2$\sqrt{2}$– 1

Answer 1

Answer: (A)

$\sqrt{2}$ – 1

Answer 2

Equation of the ellipse is

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1Equation of the parabola with focus S (ae, 0) and directrix x + ae = 0 is

y² = 4aex. Now length of the 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,

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

Ž 1 – e² = 2e Ž e² + 2e – 1 = 0

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

Hence e = $\sqrt{2}$– 1. as 0 < e < 1 for ellipse.

Hence (1) is correct answer.