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JEE Main2015MathematicsApplication of DerivativesActual

The normal to the curve x 2 + 2 x y - 3 y 2 = 0 , at 1 , 1

Options

  1. AMeets the curve again in the fourth quadrant
  2. BDoes not meet the curve again
  3. CMeets the curve again in the second quadrant
  4. DMeets the curve again in the third quadrant

Correct answer

A. Meets the curve again in the fourth quadrant

Step-by-step solution

Given x 2 + 2 x y - 3 y 2 = 0 x + 3 y x - y = 0 Pair of straight lines passing through the origin. ∵   x + 3 y = 0 or x - y = 0 Normal exists at 1 ,   1 which is on x - y = 0 ⇒ The slope of normal at 1 ,   1 = - 1 ∴ Equation of normal will be y - 1 = - x - 1 y - y 1 = m x - x 1 y - 1 = - x + 1 x + y = 2 Now, find the point of intersection with x + 3 y = 0 . x + y = 2 x + 3 y = 0 ---------------- - 2 y = 2 ⇒ y =-1,  x =3 ∵   3 ,   - 1 lies in the fourth qua

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