Question

P is a point on \[ \frac{x^2}{9}+\frac{y^2}{4}=1 \] as \(P(3\cos\alpha,2\sin\alpha)\). Q is a point on \[ x^2+y^2-14x+14y+82=0 \] R is a point on line \[ x+y=5 \] If the centroid of triangle \(PQR\) is \[ \left(\cos\alpha+2,\;\frac{2\sin\alpha}{3}+3\right) \] find the sum of possible ordinates of \(R\).

Answer 1

Correct Answer: 22

Concept: The centroid of triangle with vertices \(P(x_1,y_1), Q(x_2,y_2), R(x_3,y_3)\) is \[ \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) \]
Step 1: Point \(P\) \[ P(3\cos\alpha,2\sin\alpha) \]
Step 2: Point \(Q\) \[ x^2+y^2-14x+14y+82=0 \] Complete squares: \[ (x-7)^2+(y+7)^2=16 \] Thus parametric form: \[ Q(7+4\cos\theta,-7+4\sin\theta) \]
Step 3: Point \(R\) Given line \(x+y=5\) Let \[ R(5-y,y) \]
Step 4: Centroid \[ \left(\frac{3\cos\alpha+7+4\cos\theta+5-y}{3},\frac{2\sin\alpha-7+4\sin\theta+y}{3}\right) \] Given centroid: \[ \left(\cos\alpha+2,\frac{2\sin\alpha}{3}+3\right) \]
Step 5: Compare coordinates \[ \cos\theta=\frac{y-6}{4} \] \[ \sin\theta=\frac{16-y}{4} \]
Step 6: Use identity \[ \sin^2\theta+\cos^2\theta=1 \] \[ \left(\frac{y-6}{4}\right)^2+\left(\frac{16-y}{4}\right)^2=1 \] \[ (y-6)^2+(y-16)^2=16 \] \[ y^2-22y+138=0 \]
Step 7: Find sum Sum of roots: \[ y_1+y_2=22 \] Thus sum of possible ordinates of \(R\): \[ \boxed{22} \]