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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If eccentricity of the ellipse x^ 2 a^ 2 + 1 + y^ 2 a^ 2 + 2 = 1 is...

If eccentricity of the ellipse

\dfrac{x^{2}}{a^{2} + 1} + \dfrac{y^{2}}{a^{2} + 2} = 1

is

\dfrac{1}{\sqrt{6}},

then the ratio of the length of the latus rectum to the length of the major axis is

Options

A $\dfrac{5}{6}$
B $\dfrac{3}{\sqrt{6}}$
C $\dfrac{2}{3}$
D $\dfrac{2}{\sqrt{6}}$

Correct answer

A. $\dfrac{5}{6}$

Step-by-step solution

{a}^{2} + 2 > {a}^{2} + 1

<br />\left({a}^{2} + 1\right) = \left({a}^{2} + 2\right)\left(1 - {e}^{2}\right)

⇒ {a}^{2} = 4 ⇒ a = \pm 2

Hence, the required ratio is

\dfrac{{a}^{2} + 1}{{a}^{2} + 2} = \dfrac{5}{6}

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