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Ellipse question

2023 · 11 Apr · Shift 1 · Q35
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  5. /2023 · 11 Apr · Shift 1 · Q35

Ellipse question

2023 · 11 Apr · Shift 1 · Q35

JEE MainMathematicsEllipseMCQ+4 / −1
Consider ellipses Ek:kx2+k2y2=1,k=1,2,…,20\mathrm{E}_{k}: k x^{2}+k^{2} y^{2}=1, k=1,2, \ldots, 20Ek​:kx2+k2y2=1,k=1,2,…,20. Let Ck\mathrm{C}_{k}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\mathrm{E}_{k}Ek​. If rkr_{k}rk​ is the radius of the circle Ck\mathrm{C}_{k}Ck​, then the value of ∑k=1201rk2\sum_{k=1}^{20} \frac{1}{r_{k}^{2}}∑k=120​rk2​1​ is :
  1. A
    2870
  2. B
    3210
  3. C
    3320
  4. D
    3080
View written solutionFree

Correct answer: D

  1. Write the ellipse in standard form

Given

Ek:kx2+k2y2=1.E_k: kx^2+k^2y^2=1.Ek​:kx2+k2y2=1.

Divide by 111 appropriately:

x21/k+y21/k2=1.\frac{x^2}{1/k}+\frac{y^2}{1/k^2}=1.1/kx2​+1/k2y2​=1.

So the semi-axes are

a=1k,b=1k,a=\frac{1}{\sqrt{k}},\qquad b=\frac{1}{k},a=k​1​,b=k1​,

with a>ba>ba>b, hence the major axis is along the xxx-axis and minor axis along the yyy-axis.

Thus the four endpoints are

(±a,0),(0,±b).(\pm a,0),\qquad (0,\pm b).(±a,0),(0,±b).
  1. Find the four chords joining one major-axis endpoint and one minor-axis endpoint

These are the lines joining:

  • (a,0)(a,0)(a,0) to (0,b)(0,b)(0,b)
  • (a,0)(a,0)(a,0) to (0,−b)(0,-b)(0,−b)
  • (−a,0)(-a,0)(−a,0) to (0,b)(0,b)(0,b)
  • (−a,0)(-a,0)(−a,0) to (0,−b)(0,-b)(0,−b)

Their equations are

xa+yb=1,xa−yb=1,−xa+yb=1,−xa−yb=1.\frac{x}{a}+\frac{y}{b}=1, \qquad \frac{x}{a}-\frac{y}{b}=1, \qquad -\frac{x}{a}+\frac{y}{b}=1, \qquad -\frac{x}{a}-\frac{y}{b}=1.ax​+by​=1,ax​−by​=1,−ax​+by​=1,−ax​−by​=1.

Equivalently,

±xa±yb=1.\pm \frac{x}{a} \pm \frac{y}{b}=1.±ax​±by​=1.

These four lines are symmetric about the origin, so the circle touching all four must be centered at the origin.

  1. Radius of the circle touching these four lines

The radius is the perpendicular distance from the origin to any one of these lines. Take

xa+yb=1.\frac{x}{a}+\frac{y}{b}=1.ax​+by​=1.

In standard form:

bx+ay−ab=0.bx+ay-ab=0.bx+ay−ab=0.

Distance from (0,0)(0,0)(0,0) is

rk=∣−ab∣a2+b2=aba2+b2.r_k=\frac{|{-ab}|}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{a^2+b^2}}.rk​=a2+b2​∣−ab∣​=a2+b2​ab​.

Hence

1rk2=a2+b2a2b2=1a2+1b2.\frac{1}{r_k^2}=\frac{a^2+b^2}{a^2b^2}=\frac{1}{a^2}+\frac{1}{b^2}.rk2​1​=a2b2a2+b2​=a21​+b21​.

Now

a2=1k,b2=1k2.a^2=\frac{1}{k},\qquad b^2=\frac{1}{k^2}.a2=k1​,b2=k21​.

Therefore

1rk2=k+k2.\frac{1}{r_k^2}=k+k^2.rk2​1​=k+k2.
  1. Compute the sum

We need

∑k=1201rk2=∑k=120(k+k2)=∑k=120k+∑k=120k2.\sum_{k=1}^{20}\frac{1}{r_k^2}=\sum_{k=1}^{20}(k+k^2) =\sum_{k=1}^{20}k+\sum_{k=1}^{20}k^2.k=1∑20​rk2​1​=k=1∑20​(k+k2)=k=1∑20​k+k=1∑20​k2.

Use formulas:

∑k=120k=20⋅212=210,\sum_{k=1}^{20}k=\frac{20\cdot 21}{2}=210,k=1∑20​k=220⋅21​=210, ∑k=120k2=20⋅21⋅416=2870.\sum_{k=1}^{20}k^2=\frac{20\cdot 21\cdot 41}{6}=2870.k=1∑20​k2=620⋅21⋅41​=2870.

So

∑k=1201rk2=210+2870=3080.\sum_{k=1}^{20}\frac{1}{r_k^2}=210+2870=3080.k=1∑20​rk2​1​=210+2870=3080.
  1. Match with the options
308030803080

corresponds to Option D.

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