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JEE Main202227 Jul 2022Evening ShiftMathematicsDifferentiationActual

For the curve C : x 2 + y 2 - 3 + x 2 - y 2 - 1 5 = 0 , the value of 3 y ' - y 3 y ' ' , at the point α , α , α > 0 , on C , is equal to ________.

Correct answer

0

Step-by-step solution

Given, α , α lies on C : x 2 + y 2 - 3 + x 2 - y 2 - 1 5 = 0 So on putting α , α we get, 2 α 2 - 3 - 1 5 = 0 ⇒     α = 2 Now, differentiating the curve C we get, 2 x + 2 y · y ' + 5 x 2 - y 2 - 1 4 2 x - 2 y y ' = 0             ⋯ 1 At 2 , 2 2 + 2 y ' + 5 - 1 4 2 - 2 y ' = 0 ⇒ y ' = 3 2             ⋯ 2 Again, Diff. 1 w.r.t. x we get, 1 + y ' 2 + y y y ' ' +

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