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NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Ellipse

Let there are exactly two points on the ellipse x^ 2 a^ 2 + y^ 2 b^...

Let there are exactly two points on the ellipse

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

whose distances from

\left(0,0\right)

are equal to

\sqrt{\dfrac{a^{2}}{2} + b^{2}} .

Then, the eccentricity of the ellipse is equal to

Options

A $\dfrac{1}{2}$
B $\dfrac{1}{2\sqrt{2}}$
C $\dfrac{1}{\sqrt{2}}$
DNone of these

Correct answer

C. $\dfrac{1}{\sqrt{2}}$

Step-by-step solution

Since,&

\sqrt{\dfrac{a^{2}}{2} + b^{2}} > \left|b\right|

Hence, major axis

= 2a

& minor axis

= 2b

and

\sqrt{\dfrac{a^{2}}{2} + b^{2}} = \left|a\right| ⇒ \dfrac{a^{2}}{2} + b^{2} = a^{2}

⇒ b^{2} = \dfrac{a^{2}}{2} ⇒ \dfrac{b^{2}}{a^{2}} = \dfrac{1}{2}

⇒ 1 - e^{2} = \dfrac{1}{2} ⇒ e^{2} = \dfrac{1}{2} = e = \dfrac{1}{\sqrt{2}}

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