JEE Main20264 April 2026Evening ShiftMathematicsStraight LinesActual
Let P(3 , 2 ) , 0 , be a point on the ellipse x^2 9 + 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 (2+ , 3+ 2 3 ) . Then the sum of the ordinates of all possible points R is:
Options
- A6
- B2
- C4
- D8
Correct answer
D. 8
Step-by-step solution
Let the coordinates of the points be P(3 , 2 ) , Q(x_Q, y_Q) , and R(x_R, y_R) . The centroid of PQR is given by: ( 3 + x_Q + x_R 3 , 2 + y_Q + y_R 3 ) We are given that the centroid is (2+ , 3+ 2 3 ) . Equating the coordinates, we get: 3 + x_Q + x_R 3 = 2 + x_Q + x_R = 6 2 + y_Q + y_R 3 = 3 + 2 3 y_Q + y_R = 9 Since the point R lies on the line x+y=5 , we have x_R = 5 - y_R . Substituting x_R into the equation for x_Q : x_Q = 6 - x_R = 6 - (5 - y_R) = y_R + 1 And we already have y_Q = 9 - y_R . The point Q(x_Q, y_