Question

Let \(P(3\cos\alpha, 2\sin\alpha), \alpha \neq 0\), be a point on the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). \(Q\) be a point on the circle \(x^2 + y^2 - 14x - 14y + 82 = 0\) and \(R\) be a point on the line \(x + y = 5\) such that the centroid of the triangle \(PQR\) is \(\left( 2 + \cos\alpha, 3 + \frac{2}{3}\sin\alpha \right)\). Then the sum of the ordinates of all possible points \(R\) is:

• A. 6
• B. 2
• C. 4
• D. 8

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use the centroid formula to relate the coordinates of \(P, Q, R\). Represent \(Q\) in parametric form using the circle equation. Substitute the resulting coordinates of \(R\) into the line equation.

Step 2: Key Formula or Approach:
1. Centroid \(G = (\frac{\sum x_i}{3}, \frac{\sum y_i}{3})\).
2. Circle in standard form: \((x - 7)^2 + (y - 7)^2 = 16 \implies Q = (7 + 4\cos\theta, 7 + 4\sin\theta)\).

Step 3: Detailed Explanation:
Let \(R = (x_R, y_R)\).
\(x_G = \frac{3\cos\alpha + 7 + 4\cos\theta + x_R}{3} = 2 + \cos\alpha \implies 7 + 4\cos\theta + x_R = 6 \implies x_R = -1 - 4\cos\theta\).
\(y_G = \frac{2\sin\alpha + 7 + 4\sin\theta + y_R}{3} = 3 + \frac{2}{3}\sin\alpha \implies 7 + 4\sin\theta + y_R = 9 \implies y_R = 2 - 4\sin\theta\).
Point \(R\) lies on \(x + y = 5 \implies -1 - 4\cos\theta + 2 - 4\sin\theta = 5 \implies -4(\cos\theta + \sin\theta) = 4 \implies \cos\theta + \sin\theta = -1\).
Possible pairs \((\cos\theta, \sin\theta)\) are \((-1, 0)\) and \((0, -1)\).
For \(\sin\theta = 0, y_R = 2\).
For \(\sin\theta = -1, y_R = 2 - 4(-1) = 6\).
Sum of ordinates \(= 2 + 6 = 8\).

Step 4: Final Answer:
The sum is 8.