Mathematics - Circle Question with Solution | TestHub

Given,

$E_k: k{x}^2+k^2{y}^2=1$

$\Rightarrow E_k :\frac{x^2}{{\left(\frac{1}{\sqrt{k}}\right)}^2}+\frac{y^2}{{\left(\frac{1}{k}\right)}^2}=1$

Now equation of the chord joining the points $\left(\frac{1}{\sqrt{k}},0\right) \& \left(0,\frac{1}{k}\right)$ will be,

$L_k:\frac{x}{\left(\frac{1}{\sqrt{k}}\right)}+\frac{y}{\left(\frac{1}{k}\right)}=1$

$\Rightarrow \sqrt{k}x+ky-1=0$

Now $r_k=$ Perpendicular distance of $L_k$ from $(0,0)$ we get,

$r_k=\left|\frac{-1}{\sqrt{k+k^2}}\right|$

$\Rightarrow r_k^2=\frac{1}{k+k^2}$

Now putting the value of ${r_k}^2$ in $\sum _{k=1}^{20}\frac{1}{r_k^2}$ we get,

$\sum _{k=1}^{20}\frac{1}{r_k^2}=\sum _{k=1}^{20}k+k^2=\frac{20\times 21}{2}+\frac{20\times 21\times 41}{6}$

$=210+2870$$=3080$