JEE Main 2026 — Mathematics Coordinate Geometry

2026integerhard

Let $A,B$ be points on the two half-lines $x-\sqrt{3}|y|=\alpha$ , $\alpha>0$ at a distance of $\alpha$ from their point of intersection $P$ . The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$ . If $PQ=\dfrac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle PAB$ , then $\dfrac{\alpha^2}{R}$ is equal to ______

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

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Solution

The given equations of the half-lines are $x - \sqrt{3}y = \alpha$ for $y \ge 0$ and $x + \sqrt{3}y = \alpha$ for $y \le 0$ .

The point of intersection $P$ is obtained by setting $y = 0$ , which gives $x = \alpha$ . Thus, $P \equiv (\alpha, 0)$ .

The angle bisector of these half-lines is the x-axis ( $y = 0$ ).

The slope of the first half-line is $\dfrac{1}{\sqrt{3}}$ , so it makes an angle of $30^\circ$ with the positive x-axis. The second half-line has a slope of $-\dfrac{1}{\sqrt{3}}$ , making an angle of $-30^\circ$ with the positive x-axis.

Since $A$ and $B$ are at a distance $\alpha$ from $P$ , their coordinates can be written using the parametric form of a straight line:

$A \equiv (\alpha + \alpha \cos 30^\circ, 0 + \alpha \sin 30^\circ) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, \dfrac{\alpha}{2}\right)$

$B \equiv (\alpha + \alpha \cos(-30^\circ), 0 + \alpha \sin(-30^\circ)) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, -\dfrac{\alpha}{2}\right)$

The line segment $AB$ is vertical and intersects the angle bisector (x-axis) at $Q$ . Thus, the coordinates of $Q$ are $\left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, 0\right)$ .

The distance $PQ$ is the difference in their x-coordinates:

$PQ = \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}\right) - \alpha = \dfrac{\alpha\sqrt{3}}{2}$

Given $PQ = \dfrac{9}{2}$ , we have:

$\dfrac{\alpha\sqrt{3}}{2} = \dfrac{9}{2} \Rightarrow \alpha\sqrt{3} = 9 \Rightarrow \alpha = 3\sqrt{3}$

In $\triangle PAB$ , $PA = PB = \alpha$ and $\angle APB = 30^\circ - (-30^\circ) = 60^\circ$ . Therefore, $\triangle PAB$ is an equilateral triangle with side length $\alpha$ .

The circumradius $R$ of an equilateral triangle of side $\alpha$ is given by $R = \dfrac{\alpha}{\sqrt{3}}$ .

We need to find the value of $\dfrac{\alpha^2}{R}$ :

$\dfrac{\alpha^2}{R} = \dfrac{\alpha^2}{\dfrac{\alpha}{\sqrt{3}}} = \alpha\sqrt{3}$

Since we already found $\alpha\sqrt{3} = 9$ , we get:

$\dfrac{\alpha^2}{R} = 9$

Answer: $9$

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Solution

The given equations of the half-lines are $x - \sqrt{3}y = \alpha$ for $y \ge 0$ and $x + \sqrt{3}y = \alpha$ for $y \le 0$ .

The point of intersection $P$ is obtained by setting $y = 0$ , which gives $x = \alpha$ . Thus, $P \equiv (\alpha, 0)$ .

The angle bisector of these half-lines is the x-axis ( $y = 0$ ).

Since $A$ and $B$ are at a distance $\alpha$ from $P$ , their coordinates can be written using the parametric form of a straight line:

$A \equiv (\alpha + \alpha \cos 30^\circ, 0 + \alpha \sin 30^\circ) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, \dfrac{\alpha}{2}\right)$

$B \equiv (\alpha + \alpha \cos(-30^\circ), 0 + \alpha \sin(-30^\circ)) \equiv \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, -\dfrac{\alpha}{2}\right)$

The line segment $AB$ is vertical and intersects the angle bisector (x-axis) at $Q$ . Thus, the coordinates of $Q$ are $\left(\alpha + \dfrac{\alpha\sqrt{3}}{2}, 0\right)$ .

The distance $PQ$ is the difference in their x-coordinates:

$PQ = \left(\alpha + \dfrac{\alpha\sqrt{3}}{2}\right) - \alpha = \dfrac{\alpha\sqrt{3}}{2}$

Given $PQ = \dfrac{9}{2}$ , we have:

$\dfrac{\alpha\sqrt{3}}{2} = \dfrac{9}{2} \Rightarrow \alpha\sqrt{3} = 9 \Rightarrow \alpha = 3\sqrt{3}$

In $\triangle PAB$ , $PA = PB = \alpha$ and $\angle APB = 30^\circ - (-30^\circ) = 60^\circ$ . Therefore, $\triangle PAB$ is an equilateral triangle with side length $\alpha$ .

The circumradius $R$ of an equilateral triangle of side $\alpha$ is given by $R = \dfrac{\alpha}{\sqrt{3}}$ .

We need to find the value of $\dfrac{\alpha^2}{R}$ :

$\dfrac{\alpha^2}{R} = \dfrac{\alpha^2}{\dfrac{\alpha}{\sqrt{3}}} = \alpha\sqrt{3}$

Since we already found $\alpha\sqrt{3} = 9$ , we get:

$\dfrac{\alpha^2}{R} = 9$

Answer: $9$

JEE Main 2026 — Mathematics Coordinate Geometry

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JEE Main 2026 — Mathematics Coordinate Geometry

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