Q57 / 90 · JEE Main 2023 (11 Apr Shift 1) · Mathematics

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

  • Correct answer
    A
    3080
  • B
    2870
  • C
    3210
  • D
    3320

Solution

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

Hence, the correct answer is option: 'A'.

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