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NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Ellipse

Let the line y = mx and the ellipse 2x^ 2 + y^ 2 = 1 intersect at a...

Let the line

y = mx and the ellipse

2x^{2} + y^{2} = 1

intersect at a point

P

in the first quadrant. If the normal to this ellipse at

P

meets the co-ordinate axes at \left(- \dfrac{1}{3\sqrt{2}}, 0\right)

and \left(0, \beta\right), then

\beta

is equal to

Options

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

Correct answer

D. $\dfrac{\sqrt{2}}{3}$

Step-by-step solution

Let

P

be \left({x}_{1}, {y}_{1}\right)

<br />Equation of normal at

P

is

\dfrac{x}{2{x}_{1}} - \dfrac{y}{{y}_{1}} = - \dfrac{1}{2}

<br />It passes through \left(- \dfrac{1}{3\sqrt{2}}, 0\right) ⇒ \dfrac{- 1}{6\sqrt{2}{x}_{1}} = - \dfrac{1}{2} ⇒ {x}_{1} = \dfrac{1}{3\sqrt{2}}

So,~

{y}_{1} = \dfrac{2\sqrt{2}}{3}

(as

P

lies in Ist quadrant)
So,

\beta = \dfrac{{y}_{1}}{2} = \dfrac{\sqrt{2}}{3}

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