HyperbolamediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Rectangular Hyperbola xy=12: Area of Triangle OPQ = 7/2 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let OO be the origin, and PP and QQ be two points on the rectangular hyperbola xy=12xy=12 such that the mid point of the line segment PQPQ is (12,−12)\left(\dfrac12,-\dfrac12\right). Then the area of the triangle OPQOPQ equals
A32\dfrac32
B52\dfrac52
C72\dfrac72correct
D92\dfrac92
Solution
Step 1: For xy=12xy=12, the chord with midpoint (x1,y1)(x_1,y_1) has equation xx1+yy1=2\dfrac{x}{x_1}+\dfrac{y}{y_1}=2. With midpoint (12,−12)\left(\tfrac12,-\tfrac12\right):
x1/2+y−1/2=2 ⇒ 2x−2y=2 ⇒ x−y=1.\frac{x}{1/2}+\frac{y}{-1/2}=2\ \Rightarrow\ 2x-2y=2\ \Rightarrow\ x-y=1.
 Step 2: Solve x−y=1x-y=1 with xy=12xy=12: substitute x=y+1x=y+1 into xy=12xy=12:
(y+1)y=12 ⇒ y2+y−12=0 ⇒ y=3 or y=−4.(y+1)y=12\ \Rightarrow\ y^2+y-12=0\ \Rightarrow\ y=3\ \text{or}\ y=-4.
 So P(−3,−4)P(-3,-4) and Q(4,3)Q(4,3). Step 3: Area of △OPQ\triangle OPQ with OO at origin:
Area=12∣xPyQ−xQyP∣=12∣(−3)(3)−(4)(−4)∣=12∣−9+16∣=72.\text{Area}=\frac12\left|x_P y_Q-x_Q y_P\right|=\frac12\left|(-3)(3)-(4)(-4)\right|=\frac12\left|-9+16\right|=\frac72.
 Correct answer: (3)
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