JEE Main 2014MathematicsStraight LinesHardMCQ

JEE Main 2014 — Straight Lines Question with Solution

JEE Main 2014 (09 Apr Online)

Question

Let a and b be any two numbers satisfying 1a2+1b2=14. Then, the foot of perpendicular from the origin on the variable line xa+yb=1 lies on :

Choose an option

Show full solutionCorrect option: A
Correct answer
AA circle of radius =2

Step-by-step explanation

The given line is bx+ay=ab.

Foot of perpendicular from origin

xb=ya=--aba2+b2

x=ab2a2+b2, y=a2ba2+b2

x2+y2=a2b2a2+b2a2+b22

x2+y2=a2b2a2+b2

given 1a2+1b2=14

 a2+b2a2b2=14

⇒  x2+y2=4, which is equation of circle.
 

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About this question

This is a previous-year question from JEE Main 2014, covering the Straight Lines 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.