QJEEM23Ma0cfd: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0cfd Solution: Consider ellipses E k : k x 2 + k 2 y 2 = 1 , k = 1 , 2 , … ,…

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QJEEM23Ma0cfd: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0cfd Solution: Consider ellipses E k : k x 2 + k 2 y 2 = 1 , k = 1 , 2 , … ,…

Consider ellipses $E_{k} : k x^{2} + k^{2} y^{2} = 1, k = 1, 2, \dots, 20$ . Let $C_{k}$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $E_{k}$ . If $r_{k}$ is the radius of the circle $C_{k}$ , then the value of $\sum_{k = 1}^{20} \frac{1}{r_{k}^{2}}$ is :

Multiple Choice Options

A 2870
B 3210
C 3320
D 3080

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Accepted Answer

We have,

E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20

\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1

JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Ellipse Question 20 English Explanation

Equation of

A B

is

\frac{x}{\frac{1}{\sqrt{K}}}+\frac{y}{\frac{1}{K}}=1

or

\sqrt{K} x+K y=1

r_K

is the radius of circle

C_K

,

So,

r_K=

perpendicular distance from

(0,0)

to

\mathrm{AB}

\begin{aligned} r_{K} & = \frac{| 0 + 0 - 1 |}{\sqrt{(\sqrt{K})^{2} + K^{2}}} \\ = \frac{1}{\sqrt{K + K^{2}}} \\ \frac{1}{r_{K}^{2}} & = K + K^{2} \end{aligned}

\begin{aligned} \sum_{K = 1}^{20} \frac{1}{r_{K}^{2}} & = \sum_{K = 1}^{20} K + \sum_{K = 1}^{20} K^{2} \\ = \frac{20 \times 21}{2} + \frac{20 \times 21 \times 41}{6} \\ = 210 + 2870 = 3080 \end{aligned}

QJEEM23Ma0cfd: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0cfd Solution: Consider ellipses E k : k x 2 + k 2 y 2 = 1 , k = 1 , 2 , … ,…

Multiple Choice Options

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