Let $P(3cos\alpha,2sin\alpha),\alpha
eq 0$, be a point on the ellipse $\frac{x^{2}}{9} + \frac{y^{

Question

Let $P(3cos\alpha,2sin\alpha),\alpha 
eq 0$, 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 $\left( 2 + cos\alpha,3 + \frac{2}{3}sin\alpha ight)$. Then the sum of the ordinates of all possible points R is:

Accepted Answer

Sol. [Key Concept: Equate centroid coordinates; eliminating $\alpha$ leaves a circle in R's ordinate, so sum its roots.]Centroid :$${\left( \frac{3cos\alpha + 7 + 4cos	heta + 5 - y}{3},\frac{2sin\alpha + 7 + 4sin	heta + y}{3} ight) = \left( cos\alpha + 2,\frac{2}{3}sin\alpha + 3 ight) }{(7 + 4cos	heta,7 + 4sin	heta)}$$On comparison$${cos\alpha = \frac{y - 6}{4}\& sin\alpha = \frac{2 - y}{4} }{\sin^{2}\alpha + \cos^{2}\alpha = 1 \Rightarrow (y - 6)^{2} + (y - 2)^{2} = 16 }{y^{2} - 8y + 12 = 0 }$$⇒ Sum of ordinates $= 8$

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