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Mathematics · Hyperbola

JEE Main 2026 — 23 January, Evening Shift — Question 9

Let PQ be a chord of the hyperbola x24−y2 b2=1\frac{\mathrm{x}^{2}}{4}-\frac{\mathrm{y}^{2}}{\mathrm{~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 :

  1. Option A:

    232 \sqrt{3}

  2. Option B:

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

    Correct
  3. Option C:

    115\frac{11}{5}

  4. Option D:

    95\frac{9}{5}

Answer: B

Step-by-step solution

e=1+b4=3⇒ b=8\mathrm{e}=\sqrt{1+\frac{\mathrm{b}}{4}}=\sqrt{3} \quad \Rightarrow \mathrm{~b}=8

∴ Hyperbola x24−y28=1\frac{\mathrm{x}^{2}}{4}-\frac{\mathrm{y}^{2}}{8}=1

PMOM=tan⁡30∘\frac{\mathrm{PM}}{\mathrm{OM}}=\tan 30^{\circ}

⇒22tan⁡θ2sec⁡θ=13⇒sin⁡θ=16\Rightarrow \frac{2 \sqrt{2} \tan \theta}{2 \sec \theta}=\frac{1}{\sqrt{3}} \Rightarrow \sin \theta=\frac{1}{\sqrt{6}}

Area =2×12×OM×MP=2 \times \frac{1}{2} \times \mathrm{OM} \times \mathrm{MP}

=2sec⁡θ×22tan⁡θ=2 \sec \theta \times 2 \sqrt{2} \tan \theta

=42sin⁡θcos⁡2θ=42×16×(1−16)=835=4 \sqrt{2} \frac{\sin \theta}{\cos ^{2} \theta}=4 \sqrt{2} \times \frac{1}{\sqrt{6} \times\left(1-\frac{1}{6}\right)}=\frac{8 \sqrt{3}}{5}.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Evening Shift
Subject
Mathematics
Chapter
Hyperbola
Topic
Introduction to Hyperbola