JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Ellipse

Let

P (3 cos α, 2 sin α), α \neq = 0

, be a point on the ellipse

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1, Q

be a point on the circle

x^{2} + y^{2} - 14 x - 14 y + 82 = 0

and

R

be a point on the line

x + y = 5

such that the centroid of the triangle PQR is

(2 + cos α, 3 + \frac{2}{3} sin α)

. Then the sum of the ordinates of all possible points R is:

Answer: (D)

Solution

Write

Q = (7 + 4 cos θ, 7 + 4 sin θ)

on the circle of centre

(7, 7)

, radius

4

, and

R = (5 - y, y)

x + y = 5

. The centroid conditions give

\frac{3 cos α + ( 7 + 4 cos θ ) + ( 5 - y )}{3} = 2 + cos α

and

\frac{2 sin α + ( 7 + 4 sin θ ) + y}{3} = 3 + \frac{2}{3} sin α

. The

α

terms cancel, leaving

4 cos θ = y - 6

and

4 sin θ = 2 - y

. Since

cos^{2} θ + sin^{2} θ = 1

(y - 6)^{2} + (2 - y)^{2} = 16

, which reduces to

y^{2} - 8 y + 12 = 0

. The two ordinates of

R

are the roots of this quadratic, whose sum is

8

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option