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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2020 · 8 January · Morning Shift · Mathematics · Ellipse

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

Let the line y=mxy = mx and the ellipse 2x2+y2=12{x}^{2} + {y}^{2} = 1 intersect at a point PP in the first quadrant. If the normal to this ellipse at PP meets the co-ordinate axes at \left(- \dfrac{1}{3\sqrt{2}}, 0\right) and \left(0, \beta\right), then β\beta is equal to

Options

  1. A223\dfrac{2\sqrt{2}}{3}
  2. B23\dfrac{2}{\sqrt{3}}
  3. C23\dfrac{2}{3}
  4. D23\dfrac{\sqrt{2}}{3}

Correct answer

D. 23\dfrac{\sqrt{2}}{3}

Step-by-step solution

Let PP be \left({x}_{1}, {y}_{1}\right)<p>Equationofnormalat<p>Equation of normal atPisis\dfrac{x}{2{x}_{1}} - \dfrac{y}{{y}_{1}} = - \dfrac{1}{2}</p><p>Itpassesthrough</p><p>(−132,0)⇒−162x1=−12⇒x1=132</p><p>It passes through</p><p>\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, y1=223{y}_{1} = \dfrac{2\sqrt{2}}{3} (as PP lies in Ist quadrant)

So, β=y12=23\beta = \dfrac{{y}_{1}}{2} = \dfrac{\sqrt{2}}{3}

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