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

Step 1: Find the value of \( b^2 \).
For the hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] the eccentricity is given by \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] Here, \( a^2 = 4 \) and \( e = \sqrt{3} \). Hence, \[ 3 = 1 + \frac{b^2}{4} \Rightarrow b^2 = 8 \]
Step 2: Coordinates of points \( P \) and \( Q \).
Since \( PQ \) is perpendicular to the transverse axis, it is parallel to the \( y \)-axis. Let the points be \( (x, y) \) and \( (x, -y) \).
Step 3: Use equilateral triangle condition.
For \( \triangle OPQ \) to be equilateral, \[ OP = PQ \] This gives a relation between \( x \) and \( y \). Substituting into the hyperbola equation and solving gives the side length.
Step 4: Find the area.
The side of the equilateral triangle comes out to be \[ \frac{4}{\sqrt{5}} \] Hence, the area is \[ \frac{\sqrt{3}}{4} \times \left(\frac{4}{\sqrt{5}}\right)^2 = \frac{8\sqrt{3}}{5} \]