JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(P(3\cos\alpha, 2\sin\alpha)\), \(\alpha \neq 0\), be a point on the ellipse \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\), \(Q\) 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 \(\left(2+\cos\alpha, 3+\dfrac{2}{3}\sin\alpha\right)\). Then the sum of the ordinates of all possible points \(R\) is:
- A \(6\)
- B \(2\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
Let the coordinates of the points be \(P(3\cos\alpha, 2\sin\alpha)\), \(Q(x_Q, y_Q)\), and \(R(x_R, y_R)\). The centroid of \(\triangle PQR\) is given by: \(\left( \dfrac{3\cos\alpha + x_Q + x_R}{3}, \dfrac{2\sin\alpha + y_Q + y_R}{3} \right)\) We are given that the centroid is…
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