Consider ellipses $E_{k} : {kx}^{2} + k^{2} y^{2} = 1, k = 1, 2, \dots \dots 20$ .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

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3320

3080

3210

2870

answer is C.

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Detailed Solution

$\sum_{k} : \frac{x^{2}}{1 / k} + \frac{y^{2}}{1 / k^{2}} = 1$
Question Image
Equation of chord AB is $\frac{x}{1 / \sqrt{k}} + \frac{y}{1 / k} = 1 or \sqrt{kx} + ky = 1$
r_k = perpendicular distance of (0, 0) from the chord AB
$\begin{array}{l} ∵ r_{k} = \frac{| 0 + 0 - 1 |}{\sqrt{k + k^{2}}} = \frac{1}{\sqrt{k + k^{2}}} => \frac{1}{{r^{2}}_{k}} = k + k^{2} \\ => \sum_{k = 1}^{20} \frac{1}{r_{k}^{2}} = \sum_{k = 1}^{20} (k + k^{2}) = \sum_{k = 1}^{20} k + \sum_{k = 1}^{20} k^{2} \\ = \frac{20 \times 21}{2} + \frac{20 \times 21 \times 41}{6} = 3080 \end{array}$