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JEE Main202310 Apr 2023Morning ShiftMathematicsDifferential EquationsActual

The slope of tangent at any point x , y on a curve y = y x is x 2 + y 2 2 x y , x > 0 . If y 2 = 0 , then a value of y 8 is

Options

  1. A- 4 2
  2. B2 3
  3. C- 2 3
  4. D4 3

Correct answer

D. 4 3

Step-by-step solution

Given: d y d x = x 2 + y 2 2 x y Put y = v x ⇒ d y d x = v + x d v d x v + x d v d x = 1 + v 2 2 v ⇒ x d v d x = 1 - v 2 2 v ⇒ ∫ 2 v v 2 - 1 d v = - ∫ d x x ⇒ log e v 2 - 1 = log e C x ⇒ y 2 - x 2 x 2 = C x ⇒ y 2 - x 2 = C x Put x = 2 and y = 0 we get, 0 - 2 2 = 2 C ⇒ C = - 2 ⇒ y 2 = x 2 - 2 x ⇒ y 8 = 8 2 - 16 ⇒ y 8 = 48 = 4 3

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