JEE Main 2015MathematicsApplication of DerivativesMediumMCQ

JEE Main 2015 — Application of Derivatives Question with Solution

JEE Main 2015 (04 Apr)

Question

The normal to the curve x2+2xy-3y2=0, at 1, 1

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Show full solutionCorrect option: A
Correct answer
AMeets the curve again in the fourth quadrant

Step-by-step explanation

Given x2+2xy-3y2=0

x+3yx-y=0

Pair of straight lines passing through the origin.

∵ x+3y=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-y1=mx-x1

y-1=-x+1

x+y=2

Now, find the point of intersection with x+3y=0. 

   x+y=2x+3y=0
  ----------------
   -2y=2   ⇒y=-1, x=3

∵ 3, -1 lies in the fourth quadrant.

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About this question

This is a previous-year question from JEE Main 2015, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.