Question

Consider a parabola $y=\frac{x^2}{4}$ and the point $F(0,1)$. Let $A_1(x_1,y_1),A_2(x_2,y_2),A_3(x_3,y_3),...,A_k(x_k,y_k)$ are $\text{'}n\text{'}$ points on parabola such as $x_k>0$ and $\angle OF{A}_k=\frac{k\pi }{2n},(k=1,2,3,...,n)$. Then the value of $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^{n}F{A}_k$ is

• A. $\frac{2}{\pi }$
• B. $\frac{4}{\pi }$
• C. $\frac{8}{\pi }$
• D. $\frac{16}{\pi }$

Answer 1

The correct option is B $\frac{4}{\pi }$

Let $A_k=(2t,t^2)$

Slope of $F{A}_k=\frac{t^2-1}{2t-0}=\tan (\frac{\pi }{2}+{\theta }_k)$
$\Rightarrow \tan {\theta }_k=\frac{2t}{1-t^2}=\tan \left(2\varphi \right)$ (Say)
$\Rightarrow \varphi =\frac{{\theta }_k}{2}=\frac{k\pi }{4n}$ where $\tan \varphi =t$
Also
$F{A}_k=\sqrt{(t^2-1{)}^2+(2t{)}^2}\Rightarrow F{A}_k=t^2+1\Rightarrow F{A}_k=1+{\tan }^2\varphi \Rightarrow ={\sec }^2\left(\frac{k\pi }{4n}\right)$
$\therefore \underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^{n}F{A}_k=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n{\sec }^2\left(\frac{\pi }{4}\left(\frac{k}{n}\right)\right)$
$=\underset{0}{\overset{1}{\int }}{\sec }^2\left(\frac{\pi x}{4}\right)dx=\frac{4}{\pi }{\left[\tan \frac{\pi x}{4}\right]}_0^1=\frac{4}{\pi }$