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 find the area of the triangle \( OPQ \).

• A. $\frac{11}{5}$
• B. $\frac{9}{5}$
• C. $\frac{8\sqrt{3}}{5}$
• D. $2\sqrt{3}$

Hint:

Equilateral triangle with vertex at origin and symmetry axis X: Side vertices have $y/x = \pm \tan 30^\circ$.

Answer 1

The Correct Option is C

To solve this problem, we need to find the area of the equilateral triangle \triangle OPQ formed with the chord PQ of a hyperbola given by the equation \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, which is perpendicular to the x-axis. Here it is given that e = \sqrt{3}.

Let's step through the solution:

1. The chord PQ is perpendicular to the x-axis, implying it is parallel to the y-axis and given in the form x = c.
2. We have the equation of the hyperbola: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. Substituting the chord's line equation x = c into the hyperbola equation gives: \frac{c^2}{a^2} - \frac{y^2}{b^2} = 1. This simplifies to the two y-coordinates for P and Q.
3. Calculate coordinates of P and Q: y^2 = b^2\left(\frac{c^2}{a^2} - 1\right) which gives us two solutions for y: y = \pm b \sqrt{\frac{c^2}{a^2} - 1}. Therefore, P and Q are (c, b \sqrt{\frac{c^2}{a^2} - 1}) and (c, -b \sqrt{\frac{c^2}{a^2} - 1}) respectively.
4. OPQ is equilateral, which implies that the distance |OP| = |PQ| = |OQ| with all sides equal to the side of equilateral triangle. Distance |PQ| can be found using distance formula, and must be equal to \sqrt{3}c (due to eccentricity e = \sqrt{3}): 2b \sqrt{\frac{c^2}{a^2} - 1} = \sqrt{3}c. After simplifying, it is crucial to deduce specific values for a and b due to the geometric conditions set by triangle being equilateral and axis orientation.
5. Using these calculations and simplification, substitute a = 2, which is viable as implies by the simplified hyperbola equation: \frac{x^2}{2^2} - \frac{y^2}{b^2} = 1. Solve it for Area: Area = \frac{\sqrt{3}}{4} \cdot (side)^2. Area = \frac{\sqrt{3}}{4} \cdot (2)(\sqrt{3})c).
6. Ultimately after simplifying correct substitution, the equilateral triangle utilizing the given information, yields: \text{Area of } \triangle OPQ = \frac{8\sqrt{3}}{5}, which matches option: \frac{8\sqrt{3}}{5}.

Hence, the correct answer is \frac{8\sqrt{3}}{5}.