JEE Main 2026 Apr 4 Shift 2, Mathematics Q10: Quadratic Equations

JEE Main2026Apr 4, Shift 2Mathematics

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 $P QR$ is $(2 + cos α, 3 + \frac{2}{3} sin α)$ . Then the sum of the ordinates of all possible points $R$ is:

Solution

Circle: $(x - 7)^{2} + (y - 7)^{2} = 16$ , so $Q = (7 + 4 cos θ, 7 + 4 sin θ)$ . Let $R = (5 - y, y)$ .

Centroid x-coordinate: $\frac{3 cos α + 7 + 4 cos θ + 5 - y}{3} = cos α + 2$ .

Multiplying out: $3 cos α + 12 + 4 cos θ - y = 3 cos α + 6 \Rightarrow 4 cos θ = y - 6$ , so $cos θ = \frac{y - 6}{4}$ .

Centroid y-coordinate: $\frac{2 sin α + 7 + 4 sin θ + y}{3} = 3 + \frac{2}{3} sin α$ .

Multiplying out: $2 sin α + 7 + 4 sin θ + y = 9 + 2 sin α \Rightarrow 4 sin θ = 2 - y$ , so $sin θ = \frac{2 - y}{4}$ .

Use $sin^{2} θ + cos^{2} θ = 1$ :

$(y - 6)^{2} + (y - 2)^{2} = 16 \Rightarrow 2 y^{2} - 16 y + 40 = 16 \Rightarrow y^{2} - 8 y + 12 = 0$ .

Sum of roots (by Vieta) $= 8$ .

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JEE Main 2026 Apr 4 Shift 2, Mathematics Q10: Quadratic Equations

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