Maths Q10 – JEE-MAIN 2026

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:

Clue 1 of 3: Centroid Formula

The centroid of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is given by $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

🧠 Full Solution Path

Step 1: Express

x_{Q}, y_{Q}

in terms of

x_{R}, y_{R}

using the centroid formula✦ Active

Let $P = (x_{P}, y_{P}) = (3 \cos α, 2 \sin α)$ , $Q = (x_{Q}, y_{Q})$ , and $R = (x_{R}, y_{R})$ . The centroid $G = (x_{G}, y_{G}) = (2 + \cos α, 3 + \frac{2}{3} \sin α)$ . Using the centroid formula $x_{G} = \frac{x_{P} + x_{Q} + x_{R}}{3}$ and $y_{G} = \frac{y_{P} + y_{Q} + y_{R}}{3}$ :

\frac{3 \cos α + x_{Q} + x_{R}}{3} = 2 + \cos α ⟹ 3 \cos α + x_{Q} + x_{R} = 6 + 3 \cos α ⟹ x_{Q} + x_{R} = 6

\frac{2 \sin α + y_{Q} + y_{R}}{3} = 3 + \frac{2}{3} \sin α ⟹ 2 \sin α + y_{Q} + y_{R} = 9 + 2 \sin α ⟹ y_{Q} + y_{R} = 9

Thus, we have $x_{Q} = 6 - x_{R}$ and $y_{Q} = 9 - y_{R}$ .

Step 2: Substitute

x_{Q}, y_{Q}

into the circle equation and simplify○ Expand

The equation of the circle is $x^{2} + y^{2} - 14 x - 14 y + 82 = 0$ . Completing the square, we get $(x - 7)^{2} + (y - 7)^{2} = 16$ . Point Q lies on this circle, so $(x_{Q} - 7)^{2} + (y_{Q} - 7)^{2} = 16$ . Substitute the expressions for $x_{Q}$ and $y_{Q}$ from Step 1:

((6 - x_{R}) - 7)^{2} + ((9 - y_{R}) - 7)^{2} = 16

(- 1 - x_{R})^{2} + (2 - y_{R})^{2} = 16

(1 + x_{R})^{2} + (2 - y_{R})^{2} = 16

💡 Teacher's Secret Hint

Remember to correctly complete the square for the circle equation to find its center and radius.

Step 3: Solve for

y_{R}

using the line equation for R and find the sum of ordinates○ Expand

Point R lies on the line $x + y = 5$ , so $x_{R} + y_{R} = 5 ⟹ x_{R} = 5 - y_{R}$ . Substitute this into the equation from Step 2:

(1 + (5 - y_{R}))^{2} + (2 - y_{R})^{2} = 16

(6 - y_{R})^{2} + (2 - y_{R})^{2} = 16

(36 - 12 y_{R} + y_{R}^{2}) + (4 - 4 y_{R} + y_{R}^{2}) = 16

2 y_{R}^{2} - 16 y_{R} + 40 = 16

2 y_{R}^{2} - 16 y_{R} + 24 = 0

y_{R}^{2} - 8 y_{R} + 12 = 0

This is a quadratic equation for $y_{R}$ . Let the possible ordinates be $y_{R 1}$ and $y_{R 2}$ . The sum of the roots of a quadratic equation $a y^{2} + b y + c = 0$ is $- b / a$ . Therefore, the sum of the ordinates of all possible points R is $y_{R 1} + y_{R 2} = - (- 8) / 1 = 8$ .

💡 Teacher's Secret Hint

The problem asks for the sum of ordinates, which can be directly found from the sum of roots of the quadratic equation without solving for individual roots.

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Maths Question 10 – JEE-MAIN 2026

Clue 1 of 3: Centroid Formula

🧠 Full Solution Path