EllipsehardPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Centroid Condition: Sum of Ordinates of R = 8 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let P(3cos⁡α,2sin⁡α)P(3\cos\alpha,2\sin\alpha), α≠0\alpha\ne0, be a point on the ellipse x29+y24=1\dfrac{x^2}{9}+\dfrac{y^2}{4}=1, QQ a point on the circle x2+y2−14x−14y+82=0x^2+y^2-14x-14y+82=0, and RR a point on the line x+y=5x+y=5, such that the centroid of △PQR\triangle PQR is (2+cos⁡α, 3+23sin⁡α)\left(2+\cos\alpha,\ 3+\dfrac23\sin\alpha\right). Then the sum of the ordinates of all possible points RR is
A66
B22
C44
D88correct
Solution
Step 1: The circle has centre (7,7)(7,7) and radius 49+49−82=4\sqrt{49+49-82}=4, so Q=(7+4cos⁡θ, 7+4sin⁡θ)Q=(7+4\cos\theta,\ 7+4\sin\theta). Let R=(5−y, y)R=(5-y,\ y) on x+y=5x+y=5. Step 2: The centroid is
(3cos⁡α+7+4cos⁡θ+5−y3, 2sin⁡α+7+4sin⁡θ+y3)=(2+cos⁡α, 3+23sin⁡α).\left(\frac{3\cos\alpha+7+4\cos\theta+5-y}{3},\ \frac{2\sin\alpha+7+4\sin\theta+y}{3}\right)=\left(2+\cos\alpha,\ 3+\frac23\sin\alpha\right).
 Step 3: Comparing coordinates eliminates θ\theta and yields
cos⁡α=y−64,sin⁡α=2−y4.\cos\alpha=\frac{y-6}{4},\qquad \sin\alpha=\frac{2-y}{4}.
 Step 4: Use sin⁡2α+cos⁡2α=1\sin^2\alpha+\cos^2\alpha=1:
(y−6)2+(2−y)2=16 ⇒ 2y2−16y+40=16 ⇒ y2−8y+12=0.(y-6)^2+(2-y)^2=16\ \Rightarrow\ 2y^2-16y+40=16\ \Rightarrow\ y^2-8y+12=0.
 Step 5: Roots y=2y=2 and y=6y=6; sum of ordinates =2+6=8=2+6=8. Correct answer: (4)
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