Question

Consider a parabola $y^2=4x$ and $F$ be its focus. Let $A_1(x_1,y_1),A_2(x_2,y_2),A_3(x_3,y_3),...,A_n(x_n,y_n)$ be $n$ points on the parabola such that $x_k,y_k>0$ and $\angle OF{A}_k=\frac{k\pi }{2n}$. If $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^{n}F{A}_k=\frac{P}{\pi }$, then the value of $P$ is

Answer 1

Slope of $F{A}_k$ is
$\frac{y_k-0}{x_k-1}=-\tan {\theta }_k$
$\Rightarrow x_k=1-\frac{y_k}{\tan {\theta }_k}$

We know that $y_k^2=4x_k$
$\Rightarrow y_k^2=4(1-\frac{y_k}{\tan {\theta }_k})$
$\Rightarrow y_k^2+\left(4\cot {\theta }_k\right)y_k-4=0$
$\Rightarrow y_k=\frac{-4\cot {\theta }_k\pm \sqrt{16{\cot }^2{\theta }_k+16}}{2}$
$=\frac{-4\cot {\theta }_k+4cosec{\theta }_k}{2},\frac{-4\cot {\theta }_k-4cosec{\theta }_k}{2}$
$=2\left(\frac{1-\cos {\theta }_k}{\sin {\theta }_k}\right),-2\left(\frac{1+\cos {\theta }_k}{\sin {\theta }_k}\right)$

$y_k=2\tan \frac{{\theta }_k}{2},-2\cot \frac{{\theta }_k}{2}$
We know ${\theta }_k=\frac{k\pi }{2n}\le \frac{\pi }{2}$ and $(x_k,y_k)$ is in first quadrant so,
$y_k=2\tan \frac{{\theta }_k}{2}$
$\Rightarrow 4{\tan }^2\frac{{\theta }_k}{2}=4x_k$
$\Rightarrow x_k={\tan }^2\frac{{\theta }_k}{2}$
$\Rightarrow F{A}_k=\sqrt{(x_k-1{)}^2+(y_k{)}^2}\Rightarrow F{A}_k=\sqrt{(x_k-1{)}^2+4x_k}\Rightarrow F{A}_k=(x_k+1)\Rightarrow F{A}_k=1+{\tan }^2\frac{{\theta }_k}{2}\Rightarrow F{A}_k={\sec }^2\frac{{\theta }_k}{2}={\sec }^2\frac{k\pi }{4n}$

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^{n}F{A}_k={\int }_0^1{\sec }^2\left(\frac{\pi x}{4}\right)dx$
$=\frac{4}{\pi }{\left[\tan \left(\frac{\pi x}{4}\right)\right]}_0^1=\frac{4}{\pi }=\frac{P}{\pi }$