Question

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

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

Answer 1

Answer: (B)

Let $A_k=\left(2 t, t^2\right)$
$\therefore \text { Slope of } FA _{ k }=\frac{ t ^2-1}{2 t -0}=\tan \left(\frac{\pi}{2}+\theta_{ k }\right) $
$\therefore \tan \theta_{ k }=\frac{2 t }{1- t ^2}=\tan (2 \phi) \text { (Say) } $
$\Rightarrow \phi=\frac{\theta_{ k }}{2}=\frac{ k \pi}{4 n } \text { where tan } \phi= t$
Also $FA _{ k }=\sqrt{\left( t ^2-1\right)^2+(2 t )^2}= t ^2+1=1+\tan ^2 \phi=\sec ^2\left(\frac{ k \pi}{4 n }\right)$
$\therefore \underset{n \rightarrow \infty}{\text{Lim}} \frac{1}{n} \displaystyle\sum_{k=1}^n F A_k=\underset{n \rightarrow \infty}{\text{Lim}} \frac{1}{n} \displaystyle\sum_{k=1}^n \sec ^2\left(\frac{\pi}{4}\left(\frac{k}{n}\right)\right)$
$=\int\limits_0^1 \sec ^2\left(\frac{\pi x}{4}\right) d x=\frac{4}{\pi}\left[\tan \frac{\pi x}{4}\right]_0^1=\frac{4}{\pi}$