JEE Main 2023MathematicsCircleHardMCQ

JEE Main 2023 — Circle Question with Solution

JEE Main 2023 (11 Apr Shift 1)

Question

Consider ellipses Ek:kx2+k2y2=1,k=1,2,…,20. Let Ck be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse Ek. If rk is the radius of the circle Ck, then the value of ∑k=1201rk2 is

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Show full solutionCorrect option: A
Correct answer
A3080

Step-by-step explanation

Given,

Ek: kx2+k2y2=1

⇒Ek :x21k2+y21k2=1

Now equation of the chord joining the points 1k,0 & 0,1k will be,

Lk:x1k+y1k=1

⇒kx+ky-1=0

Now rk= Perpendicular distance of Lk from (0,0) we get,

rk=-1k+k2

⇒rk2=1k+k2

Now putting the value of rk2 in ∑k=1201rk2 we get,

∑k=1201rk2=∑k=120k+k2=20×212+20×21×416

=210+2870=3080

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About this question

This is a previous-year question from JEE Main 2023, covering the Circle 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.