JEE Main 2020MathematicsApplication of DerivativesMediumMCQ

JEE Main 2020 — Application of Derivatives Question with Solution

JEE Main 2020 (02 Sep Shift 1)

Question

Let Ph, k be a point on the curve y=x2+7x+2, nearest to the line, y=3x-3. Then the equation of the normal to the curve at P is

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Show full solutionCorrect option: A
Correct answer
Ax+3y+26=0

Step-by-step explanation

Tangent at Ph,k will be parallel to given line

dydxh,k= 2h+7 = 3 

⇒ h=–2
Point P lies on curve k=(-2)2-7×2+2=-8

For normal at P(-2,-8)

Slope of normal=-13(Negative inverse of slope of tangent)
Equation of normal : x+3y+26=0

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About this question

This is a previous-year question from JEE Main 2020, 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.