let curve $y=\dfrac{x^2}{4}$ and point $F(0,1)$. Let points $A_1(x_1,y_1),A_2(x_2,y_2),...,A_n(x_n,y_n)$ be n points on curve such that

Question

let curve $y=\dfrac{x^2}{4}$ and point $F(0,1)$. Let points $A_1(x_1,y_1),A_2(x_2,y_2),...,A_n(x_n,y_n)$ be n points on curve such that $x_k>0$ and $\angle{OFA_k}=\dfrac{k\pi}{2n}, (k=1,2,3....,n)$ then find $\lim_{n\to \infty}\dfrac{1}{n} \sum_{k=1}^{n} FA_k$

I am not able to get how to solve. Can anyone help. Thanks

Answer 1

Hint: Let $A(a,a^2/4)$ a point on $f$ and $\varphi=\angle OFA$. From $$\tan(\varphi)=\frac{a}{1-a^2/4}\quad\text{and}\quad \sin(\varphi)=\frac{a}{FA}$$ we get $$FA=\frac{2}{1+\cos(\varphi)}.$$ From here $$ \begin{align} \lim_{n\to\infty}\frac1n\sum_{k=1}^{n}\frac{2}{1+\cos(k\pi/2n)}&= \frac{2}{\pi}\lim_{n\to\infty}\frac{\pi}{2n}\sum_{k=1}^{n}\frac{2}{1+\cos(k\pi/2n)}\\ &=\frac{2}{\pi}\int_0^{\pi/2}\frac{2}{1+\cos(x)}\,dx\\ &=\frac{2}{\pi}\cdot2\\ &=\frac{4}{\pi}. \end{align} $$