JEE Main 2026 — Mathematics Coordinate Geometry

2026mcqmedium

Let chord PQ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points $P$ and $Q$ are in the ratio $1:2$ . If the chord PQ subtends an angle $\alpha$ at the focus of the parabola, then $\sin\alpha$ is equal to:

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{3}{5}$
B
$\dfrac{4}{5}$
C
$\dfrac{5}{13}$
D
$\dfrac{12}{13}$

Did you get this right?

Solution

Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 12x$ be $(x_1, y_1)$ and $(x_2, y_2)$ .

Given that the ordinates are in the ratio $1:2$ , we have $y_2 = 2y_1$ .

Since $P$ and $Q$ lie on the parabola, their abscissae are $x_1 = \dfrac{y_1^2}{12}$ and $x_2 = \dfrac{y_2^2}{12} = \dfrac{4y_1^2}{12} = \dfrac{y_1^2}{3}$ .

The length of the chord $PQ$ is $3\sqrt{13}$ , so $PQ^2 = 117$ .

Using the distance formula:

$(x_2 - x_1)^2 + (y_2 - y_1)^2 = 117$

$\left(\dfrac{y_1^2}{3} - \dfrac{y_1^2}{12}\right)^2 + (2y_1 - y_1)^2 = 117$

$\left(\dfrac{y_1^2}{4}\right)^2 + y_1^2 = 117$

$\dfrac{y_1^4}{16} + y_1^2 - 117 = 0$

$y_1^4 + 16y_1^2 - 1872 = 0$

$(y_1^2 + 52)(y_1^2 - 36) = 0$

Since $y_1^2 > 0$ , we get $y_1^2 = 36$ , which gives $y_1 = 6$ (taking the positive root by symmetry).

Substituting $y_1 = 6$ , we get $x_1 = \dfrac{36}{12} = 3$ . Thus, $P$ is $(3, 6)$ .

For $Q$ , $y_2 = 12$ and $x_2 = \dfrac{144}{12} = 12$ . Thus, $Q$ is $(12, 12)$ .

The focus of the parabola $y^2 = 12x$ is $S(3, 0)$ .

The vectors from the focus $S$ to points $P$ and $Q$ are:

$\overrightarrow{SP} = (3 - 3)\hat{i} + (6 - 0)\hat{j} = 6\hat{j}$

$\overrightarrow{SQ} = (12 - 3)\hat{i} + (12 - 0)\hat{j} = 9\hat{i} + 12\hat{j}$

The angle $\alpha$ subtended by $PQ$ at the focus is the angle between $\overrightarrow{SP}$ and $\overrightarrow{SQ}$ .

$\cos\alpha = \dfrac{\overrightarrow{SP} \cdot \overrightarrow{SQ}}{|\overrightarrow{SP}| |\overrightarrow{SQ}|}$

$\cos\alpha = \dfrac{0(9) + 6(12)}{6 \times \sqrt{9^2 + 12^2}} = \dfrac{72}{6 \times 15} = \dfrac{72}{90} = \dfrac{4}{5}$

Therefore, $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5}$

Answer: $\dfrac{3}{5}$

Your note

JEE Main Coordinate Geometry in other years

Coordinate Geometry 2025 Coordinate Geometry 2024 Coordinate Geometry 2023 Coordinate Geometry 2022 Coordinate Geometry 2021 Coordinate Geometry 2020 Coordinate Geometry 2019 Coordinate Geometry 2018 Coordinate Geometry 2017 Coordinate Geometry 2016 Coordinate Geometry 2015 Coordinate Geometry 2014

More JEE Main Coordinate Geometry questions

Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line $x+2 \sqrt{2} y=4$. If the co-ordinates of the vertex A are $(\alpha, \beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt{2} \beta|$ is

2026medium

In an equilateral triangle $PQR$, let the vertex $P$ be at $(3, 5)$ and the side $QR$ be along the line $x + y = 4$. If the orthocentre of the triangle $PQR$ is $(\alpha, \beta)$, then $9(\alpha + \beta)$ is equal to:

2026medium

Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \overrightarrow{\mathrm{d}}$. If $\lambda_{1}, \lambda_{2}\left(\lambda_{1}>\lambda_{2}\right)$ are the possible values of $(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$, then the equation $\mathrm{K}^{2} x^{2}+\left(\mathrm{K}^{2}-5 \mathrm{~K}+\lambda_{1}\right) x y+\left(3 \mathrm{~K}+\frac{\lambda_{2}}{2}\right) y^{2}-8 x+12 y+\lambda_{2}=0$ represents a circle, for K equal to :

2026hard

Consider the circle $C: x^2+y^2-6x-8y-11=0$. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle $x^2+y^2-\alpha x - \beta y - \gamma = 0$, then $\alpha + \beta + 2\gamma$ is equal to ________.

2026hard

Consider the parabola $P : y^2 = 4kx$ and the ellipse $E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of $P$ and $E$, be their latus rectums. If the eccentricity of $E$ is $e$, then $e^2 + 2\sqrt{2}$ is equal to _____.

2026hard

View all Coordinate Geometry questions

Keep practicing Coordinate Geometry

Work through every JEE Main Coordinate Geometry PYQ, year by year.

Browse all Mathematics topics

Solution

Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 12x$ be $(x_1, y_1)$ and $(x_2, y_2)$ .

Given that the ordinates are in the ratio $1:2$ , we have $y_2 = 2y_1$ .

Since $P$ and $Q$ lie on the parabola, their abscissae are $x_1 = \dfrac{y_1^2}{12}$ and $x_2 = \dfrac{y_2^2}{12} = \dfrac{4y_1^2}{12} = \dfrac{y_1^2}{3}$ .

The length of the chord $PQ$ is $3\sqrt{13}$ , so $PQ^2 = 117$ .

Using the distance formula:

$(x_2 - x_1)^2 + (y_2 - y_1)^2 = 117$

$\left(\dfrac{y_1^2}{3} - \dfrac{y_1^2}{12}\right)^2 + (2y_1 - y_1)^2 = 117$

$\left(\dfrac{y_1^2}{4}\right)^2 + y_1^2 = 117$

$\dfrac{y_1^4}{16} + y_1^2 - 117 = 0$

$y_1^4 + 16y_1^2 - 1872 = 0$

$(y_1^2 + 52)(y_1^2 - 36) = 0$

Since $y_1^2 > 0$ , we get $y_1^2 = 36$ , which gives $y_1 = 6$ (taking the positive root by symmetry).

Substituting $y_1 = 6$ , we get $x_1 = \dfrac{36}{12} = 3$ . Thus, $P$ is $(3, 6)$ .

For $Q$ , $y_2 = 12$ and $x_2 = \dfrac{144}{12} = 12$ . Thus, $Q$ is $(12, 12)$ .

The focus of the parabola $y^2 = 12x$ is $S(3, 0)$ .

The vectors from the focus $S$ to points $P$ and $Q$ are:

$\overrightarrow{SP} = (3 - 3)\hat{i} + (6 - 0)\hat{j} = 6\hat{j}$

$\overrightarrow{SQ} = (12 - 3)\hat{i} + (12 - 0)\hat{j} = 9\hat{i} + 12\hat{j}$

The angle $\alpha$ subtended by $PQ$ at the focus is the angle between $\overrightarrow{SP}$ and $\overrightarrow{SQ}$ .

$\cos\alpha = \dfrac{\overrightarrow{SP} \cdot \overrightarrow{SQ}}{|\overrightarrow{SP}| |\overrightarrow{SQ}|}$

$\cos\alpha = \dfrac{0(9) + 6(12)}{6 \times \sqrt{9^2 + 12^2}} = \dfrac{72}{6 \times 15} = \dfrac{72}{90} = \dfrac{4}{5}$

Therefore, $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5}$

Answer: $\dfrac{3}{5}$

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

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