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

JEE Main2026Apr 2, Shift 2Mathematics

Let $O$ be the origin, and $P$ and $Q$ be two points on the rectangular hyperbola $x y = 12$ such that the mid point of the line segment $P Q$ is $(\frac{1}{2}, - \frac{1}{2})$ . Then the area of the triangle $O P Q$ equals :

A
$\frac{3}{2}$
B
$\frac{5}{2}$
C
$\frac{7}{2}$
D
$\frac{9}{2}$

Solution

For the hyperbola $x y = 12$ , the chord with midpoint $(x_{1}, y_{1})$ satisfies $\frac{x}{x _{1}} + \frac{y}{y _{1}} = 2$ .

With midpoint $(\frac{1}{2}, - \frac{1}{2})$ , the chord equation becomes $2 x - 2 y = 2$ , i.e. $x - y = 1$ .

Solving with $x y = 12$ : substituting $y = x - 1$ gives $x^{2} - x - 12 = 0$ , with roots $x = 4$ and $x = - 3$ .

So $P = (- 3, - 4)$ and $Q = (4, 3)$ .

Area of $△ O P Q = \frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣ = \frac{1}{2} ∣ (- 3) (3) - (4) (- 4) ∣ = \frac{1}{2} ∣ - 9 + 16 ∣ = \frac{7}{2}$ .

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