JEE Main 2019MathematicsApplication of DerivativesEasyMCQ

JEE Main 2019 — Application of Derivatives Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

The shortest distance between the line y=x and the curve y2=x–2 is

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Show full solutionCorrect option: A
Correct answer
A742

Step-by-step explanation


y2=x–2
Differentiating w.r.t. x we get,
2yy'= 1⇒y'=  12y
For the shortest distance, the tangent at point P  will be parallel to the given line
⇒y'|p=12y1=1⇒y1=12
⇒x1=2+122=94 (∵y12=x1-2)
The shortest distance between the given curve & the line
= The perpendicular distance of point P from the line
=94-1212+12=742=742

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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.