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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Hyperbola

Let PQ be a chord of the hyperbola x^ 2 4 - y^ 2 b^ 2 =1 ,...

Let PQ be a chord of the hyperbola x24−y2b2=1\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1, perpendicular to the x -axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is 3\sqrt{3}, then the area of the triangle OPQ is

Options

  1. A232 \sqrt{3}
  2. B115\frac{11}{5}
  3. C95\frac{9}{5}
  4. D835\frac{8 \sqrt{3}}{5}

Correct answer

D. 835\frac{8 \sqrt{3}}{5}

Step-by-step solution

Hyperbola x24−y2b2=1\frac{x^2}{4} - \frac{y^2}{b^2} = 1 with e=3e = \sqrt{3}. e2=1+b24=3⇒b2=8e^2 = 1 + \frac{b^2}{4} = 3 \Rightarrow b^2 = 8.
Let P=(x0,y0)P = (x_0, y_0), Q=(x0,−y0)Q = (x_0, -y_0).
For equilateral △OPQ\triangle OPQ:
OP=PQ⇒x02+y02=4y02⇒x02=3y02OP = PQ \Rightarrow x_0^2 + y_0^2 = 4y_0^2 \Rightarrow x_0^2 = 3y_0^2.
From the hyperbola:
3y024−y028=1⇒5y028=1⇒y02=85\frac{3y_0^2}{4} - \frac{y_0^2}{8} = 1 \Rightarrow \frac{5y_0^2}{8} = 1 \Rightarrow y_0^2 = \frac{8}{5}.
Area =34(2y0)2=3⋅y02=835= \frac{\sqrt{3}}{4}(2y_0)^2 = \sqrt{3} \cdot y_0^2 = \frac{8\sqrt{3}}{5}.
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