JEE Main 2026 — Mathematics Coordinate Geometry

2026mcqmedium

Let $P(3\cos\alpha, 2\sin\alpha)$ , $\alpha \neq 0$ , be a point on the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ , $Q$ be a point on the circle $x^2+y^2-14x-14y+82=0$ and $R$ be a point on the line $x+y=5$ such that the centroid of the triangle $PQR$ is $\left(2+\cos\alpha, 3+\dfrac{2}{3}\sin\alpha\right)$ . Then the sum of the ordinates of all possible points $R$ is:

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$6$
B
$2$
C
$4$
D
$8$

Did you get this right?

Solution

Let the coordinates of the points be $P(3\cos\alpha, 2\sin\alpha)$ , $Q(x_Q, y_Q)$ , and $R(x_R, y_R)$ .

The centroid of $\triangle PQR$ is given by:

$\left( \dfrac{3\cos\alpha + x_Q + x_R}{3}, \dfrac{2\sin\alpha + y_Q + y_R}{3} \right)$

We are given that the centroid is $\left(2+\cos\alpha, 3+\dfrac{2}{3}\sin\alpha\right)$ . Equating the coordinates, we get:

$\dfrac{3\cos\alpha + x_Q + x_R}{3} = 2 + \cos\alpha \implies x_Q + x_R = 6$

$\dfrac{2\sin\alpha + y_Q + y_R}{3} = 3 + \dfrac{2}{3}\sin\alpha \implies y_Q + y_R = 9$

Since the point $R$ lies on the line $x+y=5$ , we have $x_R = 5 - y_R$ .

Substituting $x_R$ into the equation for $x_Q$ :

$x_Q = 6 - x_R = 6 - (5 - y_R) = y_R + 1$

And we already have $y_Q = 9 - y_R$ .

The point $Q(x_Q, y_Q)$ lies on the circle $x^2+y^2-14x-14y+82=0$ . Rewriting the circle's equation in standard form:

$(x-7)^2 + (y-7)^2 = 16$

Substitute $x_Q$ and $y_Q$ into the circle's equation:

$(y_R + 1 - 7)^2 + (9 - y_R - 7)^2 = 16$

$(y_R - 6)^2 + (2 - y_R)^2 = 16$

Expanding the terms:

y_R^2 - 12y_R + 36 + y_R^2 - 4y_R + 4 = 16

2y_R^2 - 16y_R + 40 = 16

2y_R^2 - 16y_R + 24 = 0

y_R^2 - 8y_R + 12 = 0

Factoring the quadratic equation:

$(y_R - 2)(y_R - 6) = 0$

Thus, the possible ordinates for point $R$ are $y_R = 2$ and $y_R = 6$ .

The sum of the ordinates of all possible points $R$ is $2 + 6 = 8$ .

Answer: $8$

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Solution

Let the coordinates of the points be $P(3\cos\alpha, 2\sin\alpha)$ , $Q(x_Q, y_Q)$ , and $R(x_R, y_R)$ .

The centroid of $\triangle PQR$ is given by:

$\left( \dfrac{3\cos\alpha + x_Q + x_R}{3}, \dfrac{2\sin\alpha + y_Q + y_R}{3} \right)$

We are given that the centroid is $\left(2+\cos\alpha, 3+\dfrac{2}{3}\sin\alpha\right)$ . Equating the coordinates, we get:

$\dfrac{3\cos\alpha + x_Q + x_R}{3} = 2 + \cos\alpha \implies x_Q + x_R = 6$

$\dfrac{2\sin\alpha + y_Q + y_R}{3} = 3 + \dfrac{2}{3}\sin\alpha \implies y_Q + y_R = 9$

Since the point $R$ lies on the line $x+y=5$ , we have $x_R = 5 - y_R$ .

Substituting $x_R$ into the equation for $x_Q$ :

$x_Q = 6 - x_R = 6 - (5 - y_R) = y_R + 1$

And we already have $y_Q = 9 - y_R$ .

The point $Q(x_Q, y_Q)$ lies on the circle $x^2+y^2-14x-14y+82=0$ . Rewriting the circle's equation in standard form:

$(x-7)^2 + (y-7)^2 = 16$

Substitute $x_Q$ and $y_Q$ into the circle's equation:

$(y_R + 1 - 7)^2 + (9 - y_R - 7)^2 = 16$

$(y_R - 6)^2 + (2 - y_R)^2 = 16$

Expanding the terms:

y_R^2 - 12y_R + 36 + y_R^2 - 4y_R + 4 = 16

2y_R^2 - 16y_R + 40 = 16

2y_R^2 - 16y_R + 24 = 0

y_R^2 - 8y_R + 12 = 0

Factoring the quadratic equation:

$(y_R - 2)(y_R - 6) = 0$

Thus, the possible ordinates for point $R$ are $y_R = 2$ and $y_R = 6$ .

The sum of the ordinates of all possible points $R$ is $2 + 6 = 8$ .

Answer: $8$

JEE Main 2026 — Mathematics Coordinate Geometry

Options

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Solution

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JEE Main 2026 — Mathematics Coordinate Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Coordinate Geometry in other years

More JEE Main Coordinate Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Vectors & 3D Geometry

Probability & Statistics

Trigonometry

Keep practicing Coordinate Geometry