Let PQ be a chord of the hyperbola x 2 4 - 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 then the area of the triangle OPQ is. [2026]

Question

Let PQ be a chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{b^{2}} = 1$ 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 $\sqrt{3}$ then the area of the triangle OPQ is. [2026]

Smart Achievers · Accepted Answer

(2)

$e = \sqrt{1 + \frac{b}{4}} = \sqrt{3} \Rightarrow b = 8$

$∴$ $Hyperbola: \frac{x^{2}}{4} - \frac{y^{2}}{8} = 1$

$\frac{P M}{O M} = \tan 30 °$

$\Rightarrow \frac{2 \sqrt{2} \tan θ}{2 \sec θ} = \frac{1}{\sqrt{3}} \Rightarrow \sin θ = \frac{1}{\sqrt{6}}$

$Area = 2 \times \frac{1}{2} \times O M \times M P$

$= 2 \sec θ \times 2 \sqrt{2} \tan θ$

$= 4 \sqrt{2} \frac{\sin θ}{\cos^{2} θ}$

$= 4 \sqrt{2} \times \frac{1}{\sqrt{6} (1 - \frac{1}{6})}$

$= \frac{8 \sqrt{3}}{5}$

Solution

Q.
Let PQ be a chord of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{b^{2}} = 1$ 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 $\sqrt{3}$ then the area of the triangle OPQ is. [2026]

1 $\frac{9}{5}$

2 $\frac{8 \sqrt{3}}{5}$

3 $\frac{11}{5}$

4 $2 \sqrt{3}$

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Solution

Q. Let PQ be a chord of the hyperbola x24-y2b2=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. [2026]

1 95

2 835

3 115

4 23