JEE Main 2026 — 4 April, Evening Shift — Question 34

Given ellipse: $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Point P: $(3\cos\alpha, 2\sin\alpha)$. Circle: $x^2 + y^2 - 14x - 14y + 82 = 0$. Complete squares: $(x-7)^2 + (y-7)^2 = 16$.

So center C = (7,7), radius 4.

Let Q = $(7+4\cos\theta, 7+4\sin\theta)$. Line: $x+y=5$. Let R = $(x_R, y_R)$ with $x_R + y_R = 5$. Centroid G of triangle PQR: $G = \left( \frac{3\cos\alpha + (7+4\cos\theta) + x_R}{3}, \frac{2\sin\alpha + (7+4\sin\theta) + y_R}{3} \right)$. Given $G = (2+\cos\alpha, 3+\frac{2}{3}\sin\alpha)$. Equate coordinates: $\frac{3\cos\alpha + 7+4\cos\theta + x_R}{3} = 2+\cos\alpha$

=> $3\cos\alpha + 7+4\cos\theta + x_R = 6+3\cos\alpha$

=> $x_R = -1 - 4\cos\theta$. $\frac{2\sin\alpha + 7+4\sin\theta + y_R}{3} = 3+\frac{2}{3}\sin\alpha$

=> $2\sin\alpha + 7+4\sin\theta + y_R = 9+2\sin\alpha$ => $y_R = 2 - 4\sin\theta$. Since R lies on line $x+y=5$, substitute: $(-1-4\cos\theta) + (2-4\sin\theta) = 5$

=> $1 - 4(\cos\theta+\sin\theta) = 5$

=> $\cos\theta+\sin\theta = -1$. Square: $1+\sin2\theta = 1$ => $\sin2\theta = 0$

=> $2\theta = n\pi$ => $\theta = \frac{n\pi}{2}$. For $\theta = \frac{\pi}{2}$: $\cos\theta=0, \sin\theta=1$

=> $x_R = -1, y_R = -2$ (sum = -3, not 5). So discard. For $\theta = \pi$: $\cos\theta=-1, \sin\theta=0$

=> $x_R = 3, y_R = 2$ (sum=5). Valid. For $\theta = \frac{3\pi}{2}$: $\cos\theta=0, \sin\theta=-1$

=> $x_R = -1, y_R = 6$ (sum=5). Valid. For $\theta = 0$: $\cos\theta=1, \sin\theta=0$

=> $x_R = -5, y_R = 2$ (sum=-3). Discard. Thus two possible R: (3,2) and (-1,6).

Sum of ordinates = 2+6 = 8.