JEE Main 2019MathematicsApplication of DerivativesEasyMCQ

JEE Main 2019 — Application of Derivatives Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

Given that the slope of the tangent to a curve y=y(x) at any point x,y is 2yx2. If the curve passes through the centre of the circle x2+y2-2x-2y=0, then its equation is

Choose an option

Show full solutionCorrect option: B
Correct answer
Bxloge⁡|y|=2(x-1)

Step-by-step explanation

Slope of tangent, dydx=2yx2
⇒dyy=2x2dx
Integrating both sides, we get,
ln⁡y=-2x+c
∵ it passes through centre 1, 1 of given circle,
⇒0=-2+c⇒c=2
∴ solution ln⁡y=-2x+2
⇒xln⁡y=-2+2x
⇒xln⁡y=2x-1

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About this question

This is a previous-year question from JEE Main 2019, 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.