First slide

Ellipse

Question

$If the eccentricity of the ellipse \frac{x^{2}}{a^{2} + 1} + \frac{y^{2}}{a^{2} + 2} = 1 is 1 / \sqrt{6}, then the latus rectum of the ellipse is$

Moderate

A

$\frac{5}{\sqrt{6}}$
B

$\frac{10}{\sqrt{6}}$
C

$\frac{8}{\sqrt{6}}$
D

$None of these$

Solution

$Here, a^{2} + 2 > a^{2} + 1$
$\begin{array}{ll} or a^{2} + 1 = (a^{2} + 2) (1 - e^{2}) \\ or a^{2} + 1 = (a^{2} + 2) \frac{5}{6} \end{array}$
$\begin{array}{l} or 6 a^{2} + 6 = 5 a^{2} + 10 \\ or a^{2} = 10 - 6 = 4 \\ or a = \pm 2 \end{array}$
$Latus rectum = \frac{2 (a^{2} + 1)}{\sqrt{a^{2} + 2}} = \frac{2 \times 5}{\sqrt{6}} = \frac{10}{\sqrt{6}}$

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