JEE Main2015MathematicsApplication of DerivativesActual
The normal to the curve x 2 + 2 x y - 3 y 2 = 0 , at 1 , 1
Options
- AMeets the curve again in the fourth quadrant
- BDoes not meet the curve again
- CMeets the curve again in the second quadrant
- 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