JEE Main202225 Jul 2022Evening ShiftMathematicsApplication of DerivativesActual
Let the area enclosed by the x -axis, and the tangent and normal drawn to the curve 4 x 3 - 3 x y 2 + 6 x 2 - 5 x y - 8 y 2 + 9 x + 14 = 0 at the point - 2 , 3 be A . Then 8 A is equal to _______.
Correct answer
0
Step-by-step solution
The given curve is 4 x 3 - 3 x y 2 + 6 x 2 - 5 x y - 8 y 2 + 9 x + 14 = 0 On differentiating both sides, we get 12 x 2 - 3 y 2 - 6 x y y ' + 12 x - 5 y - 5 x y ' - 16 y y ' + 9 = 0 Now, at the point - 2 , 3 48 - 27 + 36 y ' - 24 - 15 + 10 y ' - 48 y ' + 9 = 0 i.e. y ' - 2 , 3 = - 9 2 So slope of tangent   m T = - 9 2 and slope of normal m N = 2 9 i.e. equation of tangent T ≡ y - 3 = - 9 2 x + 2 ⇒ y = - 9 2 x - 6 and equation of normal N ≡ y - 3 = 2 9 x + 2 ⇒