Hyperbola 4x2−b2y2=1 with e=3. e2=1+4b2=3⇒b2=8.
Let P=(x0,y0), Q=(x0,−y0).
For equilateral △OPQ:
OP=PQ⇒x02+y02=4y02⇒x02=3y02.
From the hyperbola:
43y02−8y02=1⇒85y02=1⇒y02=58.
Area =43(2y0)2=3⋅y02=583.
Let PQ be a chord of the hyperbola 4x2−b2y2=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, then the area of the triangle OPQ is
Held on 23 Jan 2026 · Verified 6 Jul 2026.
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511
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583
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