Mathematics · Straight lines

JEE Main 2026 — 4 April, Evening Shift — Question 47

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 $PQ = \frac{9}{2}$ and R is the radius of the circumcircle of $\triangle PAB$ , then $\frac{\alpha^2}{R}$ is equal to _____

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

$\triangle \mathrm{PAB}$ is equilateral

\begin{aligned} \mathrm{PQ}=\frac{\sqrt{3} \alpha}{2}=\frac{9}{2} \therefore & \alpha=\frac{9}{\sqrt{3}} \\& \alpha=3 \sqrt{3} \end{aligned}

\begin{aligned} & \frac{\alpha}{2 \mathrm{r}}=\cos 30^{\circ} \\& \alpha=\frac{\sqrt{3}}{2} \cdot 2 \mathrm{R} \\& \alpha=\sqrt{3} \mathrm{R} \\& \mathrm{R}=3 \\& \frac{\alpha^{2}}{\mathrm{R}}=\frac{(3 \sqrt{3})^{2}}{3}=9 \end{aligned}

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Angle bisectors, concurrent lines.