Let PQ be a chord of the hyperbola x^24-y^2b^2=1, perpendicular to the x-axis, such that O…

Question

Let $PQ$ be a chord of the hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{b^2}=1$, 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 $\sqrt3$, then the area of the triangle $OPQ$ is:

Accepted Answer

$e^2=1+\dfrac{b^2}{4}=3 \Rightarrow b^2=8$, so the hyperbola is $\dfrac{x^2}{4}-\dfrac{y^2}{8}=1$.

Let $P=(x_1,y_1)$, $Q=(x_1,-y_1)$. Since $\triangle OPQ$ is equilateral, $\angle POQ=60^\circ$, so $\angle POX=30^\circ$, giving $\tan30^\circ=\dfrac{y_1}{x_1}=\dfrac{1}{\sqrt3}\Rightarrow y_1=\dfrac{x_1}{\sqrt3}$.

Substituting into the hyperbola: $\dfrac{x_1^2}{4}-\dfrac{x_1^2}{24}=1 \Rightarrow \dfrac{5x_1^2}{24}=1 \Rightarrow x_1^2=\dfrac{24}{5}$.

$$\text{Area}=\frac12\times PQ\times x_1 = \frac12(2y_1)x_1=x_1y_1=\frac{x_1^2}{\sqrt3}=\frac{24}{5\sqrt3}=\frac{8\sqrt3}{5}.$$

Solutions · 1