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JEE Main 202125 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Ellipse

If the curve x ^ 2 + 2 y ^ 2 = 2 intersects the line x + y = 1 at...

If the curve x2+2y2=2{x}^{2} + 2{y}^{2} = 2 intersects the line x+y=1x + y = 1 at two points PP and QQ, then the angle subtended by the line segment PQPQ at the origin isIf the curve x ^ 2 + 2 y ^ 2 = 2 intersects the line x + y = 1 at two points P and Q , then the angle subtended by the

Options

  1. Aπ2−tan⁡−1(13)\dfrac{\pi}{2} - \tan^{- 1}\left(\dfrac{1}{3}\right)
  2. Bπ2+tan⁡−1(13)\dfrac{\pi}{2} + \tan^{- 1}\left(\dfrac{1}{3}\right)
  3. Cπ2+tan⁡−1(14)\dfrac{\pi}{2} + \tan^{- 1}\left(\dfrac{1}{4}\right)
  4. Dπ2−tan⁡−1(14)\dfrac{\pi}{2} - \tan^{- 1}\left(\dfrac{1}{4}\right)

Correct answer

C. π2+tan⁡−1(14)\dfrac{\pi}{2} + \tan^{- 1}\left(\dfrac{1}{4}\right)

Step-by-step solution

If the curve x ^ 2 + 2 y ^ 2 = 2 intersects the line x + y = 1 at two points P and Q , then the angle subtended by the

Homogenising

x2+2y2−2(x+y)2=0{x}^{2} + 2{y}^{2} - 2{\left(x + y\right)}^{2} = 0

⇒−x2−4xy=0⇒x2+4xy=0⇒ - {x}^{2} - 4xy = 0 ⇒ {x}^{2} + 4xy = 0

Lines are x = 0 and y=−x4y = - \dfrac{x}{4}

∴∴ Angle between lines =π2+tan⁡−114= \dfrac{\pi}{2} + \tan^{- 1}\dfrac{1}{4}

option \left(3\right)

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