If the eccentricity of the ellipse x2a2+1+y2a2+2=1 be 16then length of the latusrectum of the ellipse is

e=a2+2−a2−1a2+2=16a2+2=6⇒a2=4⇒LL'=2a2b=106

If the eccentricity of the ellipse x2a2+1+y2a2+2=1 be 16then length of the latusrectum of the ellipse is

e=a2+2−a2−1a2+2=16a2+2=6⇒a2=4⇒LL'=2a2b=106

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If the eccentricity of the ellipse $\frac{x^{2}}{a^{2} + 1} + \frac{y^{2}}{a^{2} + 2} = 1$ be $\frac{1}{\sqrt{6}}$ then length of the latusrectum of the ellipse is

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

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

$\frac{5}{\sqrt{2}}$

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

answer is B.

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Detailed Solution

$e = \sqrt{\frac{a^{2} + 2 - a^{2} - 1}{a^{2} + 2}} = \frac{1}{\sqrt{6}}$

$a^{2} + 2 = 6 \Rightarrow a^{2} = 4$

$\Rightarrow L L' = \frac{2 a^{2}}{b} = \frac{10}{\sqrt{6}}$

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If the eccentricity of the ellipse x2a2+1+y2a2+2=1 be 16then length of the latusrectum of the ellipse is

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If the eccentricity of the ellipse $\frac{x^{2}}{a^{2} + 1} + \frac{y^{2}}{a^{2} + 2} = 1$ be $\frac{1}{\sqrt{6}}$ then length of the latusrectum of the ellipse is