Question

PQ is chord of hyperbola $\frac{x^2{4} - \frac{y^2}{b^2} = 1$ perpendicular to x-axis. $\triangle OPQ$ is equilateral ($e=\sqrt{3}$). Area OPQ is}

• 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

Relation $b^2 = a^2(e^2-1)$. Here $a^2=4, e^2=3$.
$b^2 = 4(2) = 8$. Hyperbola: $\frac{x^2}{4} - \frac{y^2}{8} = 1$.
PQ is vertical chord $x=k$. P is $(k, y_1)$. $\triangle OPQ$ equilateral with vertex at origin implies $\angle POX = 30^\circ$.
$y_1 / k = \tan 30^\circ = 1/\sqrt{3} \implies k^2 = 3y_1^2$.
Substitute into hyperbola: $\frac{3y_1^2}{4} - \frac{y_1^2}{8} = 1$.
$\frac{5y_1^2}{8} = 1 \implies y_1^2 = 8/5$.
Area = $\frac{\sqrt{3}}{4}(\text{Side})^2$. Side $= 2y_1$.
Area = $\sqrt{3}y_1^2 = \sqrt{3}(8/5) = \frac{8\sqrt{3}}{5}$.