Verified Solution • 2025–26 Session moderate

QJEEM21Ma24fe: JEE Main 2021 Mathematics - Class 11 JEEM21Ma24fe Solution: If the curve x 2 + 2y 2 = 2 intersects the line x + y = 1 at…

If the curve x2 + 2y2 = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is :

Multiple Choice Options

  • A π2−tan⁡−1(14){\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 4}} \right)
  • B π2+tan⁡−1(13){\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 3}} \right)
  • C π2−tan⁡−1(13){\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 3}} \right)
  • D π2+tan⁡−1(14){\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)

Step-by-Step Verified Solution

Ellipse : x2+y1=1{x \over 2} + {y \over 1} = 1

Line : x+y=1x + y = 1

JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Ellipse Question 56 English Explanation

Using homogenisation

x2+2y2=2(1)2{x^2} + 2{y^2} = 2{(1)^2}

x2+2y2=2(x+y)2{x^2} + 2{y^2} = 2{(x + y)^2}

x2+2y2=2x2+2y2+4xy{x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy

x2+4xy=0{x^2} + 4xy = 0

for ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0

tan⁡θ=∣2h2−aba+b∣\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|

tan⁡θ=∣2(2)2−01+0∣\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|

tan⁡θ=−4\tan \theta = - 4

cot⁡θ=−14\cot \theta = - {1 \over 4}

θ=cot⁡−1(−14)\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)

θ=π−cot⁡−1(14)\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)

θ=π−(π2−tan⁡−1(14))\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)

θ=π2+tan⁡−1(14)\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)

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