JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 11)
Consider ellipses $$\mathrm{E}_{k}: k x^{2}+k^{2} y^{2}=1, k=1,2, \ldots, 20$$. Let $$\mathrm{C}_{k}$$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $$\mathrm{E}_{k}$$. If $$r_{k}$$ is the radius of the circle $$\mathrm{C}_{k}$$, then the value of $$\sum_\limits{k=1}^{20} \frac{1}{r_{k}^{2}}$$ is :
2870
3210
3320
3080
Explanation
We have, $E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20$
$\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1$
_11th_April_Morning_Shift_en_11_1.png)
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 $\mathrm{AB}$
$$ \begin{aligned} r_K & =\frac{|0+0-1|}{\sqrt{(\sqrt{K})^2+K^2}} \\\\ & =\frac{1}{\sqrt{K+K^2}} \\\\ \frac{1}{r_K^2} & =K+K^2 \end{aligned} $$
$$ \begin{aligned} \sum_{K=1}^{20} \frac{1}{r_K^2} & =\sum_{K=1}^{20} K+\sum_{K=1}^{20} K^2 \\\\ & =\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\\\ & =210+2870=3080 \end{aligned} $$
$\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1$
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 $\mathrm{AB}$
$$ \begin{aligned} r_K & =\frac{|0+0-1|}{\sqrt{(\sqrt{K})^2+K^2}} \\\\ & =\frac{1}{\sqrt{K+K^2}} \\\\ \frac{1}{r_K^2} & =K+K^2 \end{aligned} $$
$$ \begin{aligned} \sum_{K=1}^{20} \frac{1}{r_K^2} & =\sum_{K=1}^{20} K+\sum_{K=1}^{20} K^2 \\\\ & =\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\\\ & =210+2870=3080 \end{aligned} $$
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