JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

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

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Solution

The half-lines $x - 3 y = α$ (upper) and $x + 3 y = α$ (lower) meet at $P = (α, 0)$ and make angles $\pm 30°$ with the x-axis, so the angle between them at $P$ is $60°$ .

The angle bisector is the x-axis. With $P A = P B = α$ and angle $A P B = 60°$ , $△ P A B$ is equilateral. $Q$ is the foot of perpendicular from $P$ to $A B$ (midpoint of $A B$ ).

$P Q = \frac{3 α}{2} = \frac{9}{2} \Rightarrow α = \frac{9}{3} = 33$ .

Circumradius of equilateral triangle of side $α$ : $R = \frac{α}{3} = \frac{3 3}{3} = 3$ .

$\frac{α ^{2}}{R} = \frac{( 3 3 ) ^{2}}{3} = \frac{27}{3} = 9$ .

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