Skip to content
dronacharya.ai
← JEE Main 2026 · 4 Apr · Shift 2
MathematicsSingle correct+4 / -1Question 10 of 75

JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 10

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, Q be a point on the circle x2+y2−14x−14y+82=0x^2+y^2-14x-14y+82=0 and R be a point on the line x+y=5x+y=5 such that the centroid of the 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 R is:

Choose one option
Answer and worked solution

Official answer: D

Centroid. Comparing with P's coordinates: Qx+Rx=6Q_x+R_x=6 and Qy+Ry=9Q_y+R_y=9.

Parametrise. With R=(t,5−t)R=(t,5-t): Q=(6−t,4+t)Q=(6-t,4+t).

Q on the circle. 2t2−4t−6=02t^2-4t-6=0, so t=3t=3 or t=−1t=-1.

Ordinates of R are 2 and 6; sum 8.

(D)