Consider ellipses E_k:kx^2+k^2y^2=1,k=1,2, ,20.

Question

Consider ellipses ${E}_{k}:k{x}^{2}+{k}^{2}{y}^{2}=1,k=1,2,\ldots ,20$. 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 $\sum _{k=1}^{20}\frac{1}{{r}_{2}^{k}}$ is :

Accepted Answer

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

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

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 $\mathrm{AB}$

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

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

Solutions · 1