Question

If \( PQ \) is a chord perpendicular to the transverse axis of \[ \frac{x^2}{4} - \frac{y^2}{b^2} = 1 \] of eccentricity \( \sqrt{3} \) such that \( \triangle OPQ \) is an equilateral triangle (where \( O \) is the origin), then the area of \( \triangle OPQ \) is:

• A. \( \dfrac{4\sqrt{3}}{5} \)
• B. \( \dfrac{2\sqrt{3}}{5} \)
• C. \( \dfrac{8\sqrt{3}}{5} \)
• D. \( \dfrac{16\sqrt{3}}{5} \)

Hint:

For conic sections, always compute unknown parameters like \( b^2 \) using the eccentricity before proceeding.

Answer 1

The Correct Option is C

To solve this problem, we need to find the area of triangle \( \triangle OPQ \), where \( PQ \) is a chord perpendicular to the transverse axis of the hyperbola given by: 

\(\frac{x^2}{4} - \frac{y^2}{b^2} = 1\).

We are also given that the eccentricity (e) of the hyperbola is \( \sqrt{3} \).

First, let's understand the properties and equations related to the hyperbola:

• The transverse axis of the hyperbola is along the x-axis because \(\frac{x^2}{4}\) is positive.

Using the formula for eccentricity \(e = \sqrt{1 + \frac{b^2}{4}}\), we have:

\(\sqrt{1 + \frac{b^2}{4}} = \sqrt{3}\)

By squaring and solving, we get:

• \(1 + \frac{b^2}{4} = 3\)
• \(\frac{b^2}{4} = 2\)
• \(b^2 = 8\)

Since \( \triangle OPQ \) is an equilateral triangle, \( OQ = PQ = OP \).

Let the coordinates of \( P \) and \( Q \) be \( (a, b) \) and \( (a, -b) \) respectively.

This is because \( PQ \) is a vertical line parallel to the y-axis, at a distance \( a \) from the origin along the x-axis.

The condition \( \triangle OPQ \) is equilateral gives:

• \(OP = OQ = PQ\)

Solving the equation:

• Squaring both sides, \(4b^2 = a^2 + b^2\)
• Rearrange to get, \(3b^2 = a^2\)

Now, the area of the equilateral triangle \( \triangle OPQ \) is given by:

• \(\frac{\sqrt{3}}{4} \times (side)^2\)

Thus, the correct answer is \(\frac{8\sqrt{3}}{5}\).