JEE Main 2026 — Mathematics Coordinate Geometry

The equation of the chord of the hyperbola $xy = 12$xy=12 with midpoint $(x_1, y_1)$(x1​,y1​) is given by $T = S_1$T=S1​.

$$\dfrac{x y_1 + y x_1}{2} - 12 = x_1 y_1 - 12$$2xy1​+yx1​​−12=x1​y1​−12

$$x y_1 + y x_1 = 2 x_1 y_1$$xy1​+yx1​=2x1​y1​

Substituting the midpoint $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$(21​,−21​):

$$x\left(-\dfrac{1}{2}\right) + y\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)$$x(−21​)+y(21​)=2(21​)(−21​)

$$-x + y = -1 \Rightarrow y = x - 1$$−x+y=−1⇒y=x−1

To find the coordinates of P and Q, substitute $y = x - 1$y=x−1 into the equation of the hyperbola $xy = 12$xy=12:

$$x(x - 1) = 12$$x(x−1)=12

$$x^2 - x - 12 = 0$$x2−x−12=0

$$(x - 4)(x + 3) = 0 \Rightarrow x = 4, -3$$(x−4)(x+3)=0⇒x=4,−3

For $x = 4$x=4, $y = 3$y=3.

For $x = -3$x=−3, $y = -4$y=−4.

Thus, the coordinates of P and Q are $(4, 3)$(4,3) and $(-3, -4)$(−3,−4).

The area of triangle OPQ with vertices $(0,0)$(0,0), $(4,3)$(4,3), and $(-3,-4)$(−3,−4) is:

$$\text{Area} = \dfrac{1}{2} |x_P y_Q - x_Q y_P|$$Area=21​∣xP​yQ​−xQ​yP​∣

$$\text{Area} = \dfrac{1}{2} |(4)(-4) - (-3)(3)|$$Area=21​∣(4)(−4)−(−3)(3)∣

$$\text{Area} = \dfrac{1}{2} |-16 + 9| = \dfrac{7}{2}$$Area=21​∣−16+9∣=27​

Answer: $\dfrac{7}{2}$27​