Question

Let $P$, $Q$ and $R$ be the three points on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Eccentric angles of $P$, $Q$ and $R$ are $\alpha$, $\alpha + \frac{2\pi}{3}$ and $\alpha + \frac{4\pi}{3}$ respectively.

• A. (A) If area of $\triangle PQR$ is $\frac{3\sqrt{3}}{4\pi}$ times the area of ellipse then '$\alpha$' can be
• B. (B) If area of triangle formed by normals at $P$, $Q$ and $R$ is zero, then '$\alpha$' can be
• C. (C) If area of triangle formed by tangents at $P$, $Q$ and $R$ is $\frac{3\sqrt{3}}{\pi}$ times the area of ellipse then '$\alpha$' can be
• D. (D) If area of triangle formed by lines joining the mid-points of chords $PQ$, $QR$ and $RP$ is maximum then '$\alpha$' can be

Answer 1

A-p, B-r, C-q, D-s

Answer 2

The problem is a matching question involving properties of points on an ellipse with specific eccentric angles.

Part (A): Area of $\triangle PQR$ The area of $\triangle PQR$ with eccentric angles $\alpha$, $\alpha + \frac{2\pi}{3}$, $\alpha + \frac{4\pi}{3}$ is $A_{\triangle PQR} = \frac{3\sqrt{3}}{4} ab$. The area of the ellipse is $A_{ellipse} = \pi ab$. The ratio is $\frac{A_{\triangle PQR}}{A_{ellipse}} = \frac{\frac{3\sqrt{3}}{4} ab}{\pi ab} = \frac{3\sqrt{3}}{4\pi}$. This ratio is constant and independent of $\alpha$. Therefore, the condition is always met. Any $\alpha$ can be a valid answer. Matching with (p) $\alpha = 0$.

Part (B): Area of triangle formed by normals at $P$, $Q$ and $R$ is zero. The normals at $P, Q, R$ are concurrent if $\alpha + (\alpha + \frac{2\pi}{3}) + (\alpha + \frac{4\pi}{3}) = n\pi$ for some integer $n$. This simplifies to $3\alpha + 2\pi = n\pi$, or $3\alpha = (n-2)\pi$. Let $m = n-2$, so $\alpha = \frac{m\pi}{3}$. Checking the options:

• (p) $\alpha = 0$: $0 = m\pi/3 \implies m=0$. Valid.
• (r) $\alpha = \pi/3$: $\pi/3 = m\pi/3 \implies m=1$. Valid. So, $\alpha$ can be $0$ or $\pi/3$. Matching with (r) $\alpha = \pi/3$.

Part (C): Area of triangle formed by tangents at $P$, $Q$ and $R$ The area of the triangle formed by tangents at points with eccentric angles $\theta_1, \theta_2, \theta_3$ is $A = ab |\tan\left(\frac{\theta_2-\theta_1}{2}\right) \tan\left(\frac{\theta_3-\theta_2}{2}\right) \tan\left(\frac{\theta_1-\theta_3}{2}\right)|$. Here, $\frac{\theta_2-\theta_1}{2} = \frac{\pi}{3}$, $\frac{\theta_3-\theta_2}{2} = \frac{\pi}{3}$, $\frac{\theta_1-\theta_3}{2} = -\frac{2\pi}{3}$. $A_{\text{tangents}} = ab |\tan(\frac{\pi}{3}) \tan(\frac{\pi}{3}) \tan(-\frac{2\pi}{3})| = ab |(\sqrt{3})(\sqrt{3})(\sqrt{3})| = 3\sqrt{3} ab$. The ratio to the ellipse area is $\frac{3\sqrt{3} ab}{\pi ab} = \frac{3\sqrt{3}}{\pi}$. This ratio is constant and independent of $\alpha$. Matching with (q) $\alpha = \pi/6$.

Part (D): Area of triangle formed by lines joining the mid-points of chords $PQ$, $QR$ and $RP$ is maximum. The triangle formed by the midpoints of chords $PQ, QR, RP$ has an area that is $1/4$ of the area of $\triangle PQR$. Area($\triangle M_{PQ}M_{QR}M_{RP}$) = $\frac{1}{4} \times \frac{3\sqrt{3}}{4} ab = \frac{3\sqrt{3}}{16} ab$. This area is constant and independent of $\alpha$. Thus, it is always maximum. Matching with (s) $\alpha = \pi/2$.

The matches are: (A) $\to$ (p), (B) $\to$ (r), (C) $\to$ (q), (D) $\to$ (s).