JEE Main 2023 — Mathematics Coordinate Geometry

Given,

$${E}_{k}:k{x}^{2}+{k}^{2}{y}^{2}=1$$Ek​:kx2+k2y2=1

$$\Rightarrow {E}_{k}:\frac{{x}^{2}}{{(\frac{1}{\sqrt{k}})}^{2}}+\frac{{y}^{2}}{{(\frac{1}{k})}^{2}}=1$$⇒Ek​:(k​1​)2x2​+(k1​)2y2​=1

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

$${L}_{k}:\frac{x}{(\frac{1}{\sqrt{k}})}+\frac{y}{(\frac{1}{k})}=1$$Lk​:(k​1​)x​+(k1​)y​=1

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

Now ${r}_{k}=$rk​= Perpendicular distance of ${L}_{k}$Lk​ from $(0,0)$(0,0) we get,

$${r}_{k}=|\frac{-1}{\sqrt{k+{k}^{2}}}|$$rk​=∣k+k2​−1​∣

$$\Rightarrow {r}_{k}^{2}=\frac{1}{k+{k}^{2}}$$⇒rk2​=k+k21​

Now putting the value of ${{r}_{k}}^{2}$rk​2 in $\displaystyle \sum _{k=1}^{20}\frac{1}{{r}_{k}^{2}}$k=1∑20​rk2​1​ we get,

$$\sum _{k=1}^{20}\frac{1}{{r}_{k}^{2}}=\sum _{k=1}^{20}k+{k}^{2}=\frac{20\times 21}{2}+\frac{20\times 21\times 41}{6}$$k=1∑20​rk2​1​=k=1∑20​k+k2=220×21​+620×21×41​

$=210+2870$=210+2870$=3080$=3080