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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 11

Let O be the origin, and P and Q be two points on the rectangular hyperbola xy=12xy=12 such that the mid point of the line segment PQ is (12,−12)\left(\dfrac12,-\dfrac12\right). Then the area of the triangle OPQ equals :

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Answer and worked solution

Official answer: C

Points. P(a,12a)P(a,\frac{12}a), Q(b,12b)Q(b,\frac{12}b) with midpoint conditions a+b=1a+b=1 and 12(a+b)ab=−1\frac{12(a+b)}{ab}=-1, so ab=−12ab=-12.

Area.
Area=12∣a⋅12b−b⋅12a∣=6∣a2−b2∣∣ab∣=∣a−b∣2.\text{Area}=\frac12\left|a\cdot\frac{12}b-b\cdot\frac{12}a\right|=6\frac{|a^2-b^2|}{|ab|}=\frac{|a-b|}2.

Value. (a−b)2=1+48=49(a-b)^2=1+48=49, so the area is 72\frac72.

(C)

Topics:HyperbolaDistance and section formulae
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