Question

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

• A. $\frac{2}{\pi }$
• B. $\frac{4}{\pi }$
• C. $\frac{8}{\pi }$
• D. none

Answer 1

Answer: (B)

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