JEE Challenger
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MathMedMain 2026 • 4th Apr - Shift 2

Sum of Ordinates of Points R Given Centroid of PQR

Let P(3cos⁡α,2sin⁡α)P(3\cos\alpha, 2\sin\alpha), α≠0\alpha \neq 0, be a point on the ellipse x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1, QQ be a point on the circle x2+y2−14x−14y+82=0x^2 + y^2 - 14x - 14y + 82 = 0 and RR be a point on the line x+y=5x + y = 5 such that the centroid of the triangle PQRPQR is (2+cos⁡α,3+23sin⁡α)\left(2 + \cos\alpha, 3 + \frac{2}{3}\sin\alpha\right). Then the sum of the ordinates of all possible points RR is:

Options

A

6

B

2

C

4

D

8

Correct

Topics & Concepts

Conic SectionsEllipseCirclesStraight Lines

Step-by-Step Solution

To find the sum of the ordinates of all possible points RR, we analyze the given information step by step.

Step 1: Simplify the equation of the circle The equation of the circle on which point QQ lies is: x2+y2−14x−14y+82=0x^2 + y^2 - 14x - 14y + 82 = 0

By completing the square, we can rewrite this as: (x−7)2−49+(y−7)2−49+82=0(x - 7)^2 - 49 + (y - 7)^2 - 49 + 82 = 0 (x−7)2+(y−7)2=16(x - 7)^2 + (y - 7)^2 = 16

Thus, Q(x2,y2)Q(x_2, y_2) lies on the circle with center (7,7)(7, 7) and radius 44.

Step 2: Relate the coordinates using the centroid formula Let the coordinates of the points be:

  • P=(3cos⁡α,2sin⁡α)P = (3\cos\alpha, 2\sin\alpha)
  • Q=(x2,y2)Q = (x_2, y_2)
  • R=(x3,y3)R = (x_3, y_3)

The centroid GG of △PQR\triangle PQR is given by: G=(3cos⁡α+x2+x33,2sin⁡α+y2+y33)G = \left(\frac{3\cos\alpha + x_2 + x_3}{3}, \frac{2\sin\alpha + y_2 + y_3}{3}\right)

We are given that the centroid is (2+cos⁡α,3+23sin⁡α)\left(2 + \cos\alpha, 3 + \frac{2}{3}\sin\alpha\right).

Equating the xx-coordinates: 3cos⁡α+x2+x33=2+cos⁡α\frac{3\cos\alpha + x_2 + x_3}{3} = 2 + \cos\alpha 3cos⁡α+x2+x3=6+3cos⁡α3\cos\alpha + x_2 + x_3 = 6 + 3\cos\alpha x2+x3=6  ⟹  x2=6−x3x_2 + x_3 = 6 \implies x_2 = 6 - x_3

Equating the yy-coordinates: 2sin⁡α+y2+y33=3+23sin⁡α\frac{2\sin\alpha + y_2 + y_3}{3} = 3 + \frac{2}{3}\sin\alpha 2sin⁡α+y2+y3=9+2sin⁡α2\sin\alpha + y_2 + y_3 = 9 + 2\sin\alpha y2+y3=9  ⟹  y2=9−y3y_2 + y_3 = 9 \implies y_2 = 9 - y_3

Step 3: Use the condition for point RR Since R(x3,y3)R(x_3, y_3) lies on the line x+y=5x + y = 5, we have: x3+y3=5  ⟹  x3=5−y3x_3 + y_3 = 5 \implies x_3 = 5 - y_3

Substitute x3=5−y3x_3 = 5 - y_3 into the expression for x2x_2: x2=6−(5−y3)=1+y3x_2 = 6 - (5 - y_3) = 1 + y_3

Thus, point QQ in terms of y3y_3 is: Q=(1+y3,9−y3)Q = (1 + y_3, 9 - y_3)

Step 4: Solve for y3y_3 using the equation of the circle Since Q(1+y3,9−y3)Q(1 + y_3, 9 - y_3) lies on the circle (x−7)2+(y−7)2=16(x - 7)^2 + (y - 7)^2 = 16: (1+y3−7)2+(9−y3−7)2=16(1 + y_3 - 7)^2 + (9 - y_3 - 7)^2 = 16 (y3−6)2+(2−y3)2=16(y_3 - 6)^2 + (2 - y_3)^2 = 16

Expanding the terms: (y32−12y3+36)+(y32−4y3+4)=16(y_3^2 - 12y_3 + 36) + (y_3^2 - 4y_3 + 4) = 16 2y32−16y3+40=162y_3^2 - 16y_3 + 40 = 16 2y32−16y3+24=02y_3^2 - 16y_3 + 24 = 0 y32−8y3+12=0y_3^2 - 8y_3 + 12 = 0

Factoring the quadratic equation: (y3−2)(y3−6)=0(y_3 - 2)(y_3 - 6) = 0

Thus, the possible ordinates (y3y_3) of point RR are y3=2y_3 = 2 and y3=6y_3 = 6.

Step 5: Calculate the sum of the ordinates Sum of ordinates=2+6=8\text{Sum of ordinates} = 2 + 6 = 8

Correct Option: D

Sum of Ordinates of Points R Given Centroid of PQR | Mathematics PYQ Solution - JEE Challenger