Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
Consider a parabola $4y = x^2$ and point $B(0,1)$. Let $A_1\left(x_1, y_1\right), A_2\left(x_2, y_2\right), \ldots\ldots\ldots\ldots, A_n\left(x_n, y_n\right)$ are $n$ points on the parabola such that $x_r > 0$ and $\angle OBA_r = \frac{r\pi}{2n}$ $(r = 1, 2, \ldots\ldots\ldots, n)$ then $\pi\left(\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} BA_r\right)$ is equal to ______.
Step-by-Step Solution
Key Concept: Converting a discrete sum of slopes into a Riemann integral using parametric representation and trigonometric identities.
Let $A_t = (2t, t^2)$ on the parabola. The slope of $BA_t$ is $BA_t = \frac{t^2-1}{2t} = \tan\left(\frac{\pi}{2} + \theta_t\right)$. Using $\tan(\theta_t) = \frac{2t}{1-t^2} = \tan(2\phi)$, we get $\phi = \frac{\theta_t}{2} - \frac{\pi}{4n}$. The limit $\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}BA_r = \int_0^{\pi/4}\sec^2\left(\frac{\pi x}{4}\right)dx = \frac{4}{\pi}$.
Correct Answer: 4