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Maths Question 11 – JEE-MAIN 2026

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Let O be the origin, and P and Q be two points on the rectangular hyperbola xy=12 such that the mid point of the line segment PQ is (12,12). Then the area of the triangle OPQ equals :

Recall the formula for the equation of a chord of a hyperbola xy=c2 when its midpoint (h,k) is given.

Step 1: Determine the equation of the chord PQ✦ Active

The midpoint of the line segment PQ is M(12,12). For a hyperbola xy=c2, the equation of the chord with midpoint (h,k) is xk+yh=2hk. Here, h=12 and k=12. The equation of the chord PQ is x(12)+y(12)=2(12)(12), which simplifies to x+y=1.

Note: The original problem statement xy=12 with x+y=1 leads to complex roots for P and Q. Given the real-valued options, it is highly probable that the intended hyperbola equation was xy=−12. We will proceed with this corrected assumption to find a valid real solution.

Step 2: Find the coordinates of points P and Q○ Expand

Solve the system of equations for the corrected hyperbola and the chord:

xy=−12(corrected hyperbola) x+y=1⟹y=1−x

Substitute y into the hyperbola equation:

x(1−x)=−12 x−x2=−12 x2−x−12=0

Factoring the quadratic equation:

(x−4)(x+3)=0

The roots are x1=4 and x2=−3. Corresponding y-coordinates are:

For x1=4,y1=1−4=−3⟹P=(4,−3) For x2=−3,y2=1−(−3)=4⟹Q=(−3,4)
Step 3: Calculate the area of triangle OPQ○ Expand

The vertices of the triangle are O(0,0), P(4,−3), and Q(−3,4). The area of triangle OPQ is given by the formula 12|x1y2−x2y1|.

Area=12|(4)(4)−(−3)(−3)| Area=12|16−9| Area=12|7| Area=72
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