Let PQ be a chord of the hyperbola (x^2)/(4)-(y^2)/(b^2)=1, perpendicular to the x -axis s…

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 the a...

Accepted Answer

PQ is a chord perpendicular to x -axis of hyperbola $\frac{{x}^{2}}{4}-\frac{{y}^{2}}{{b}^{2}}=1$

$\begin{gathered} e=\sqrt{3} \ e=\sqrt{1+\frac{{b}^{2}}{4}}=\sqrt{3} \ 1+\frac{{b}^{2}}{4}=3 \end{gathered}$

$\frac{{b}^{2}}{4}=2\implies {b}^{2}=8$

equation become $\frac{{x}^{2}}{4}-\frac{{y}^{2}}{8}=1$ - (1)

let coordinate of point $P({x}_{1},{y}_{1})$

PQ is perpendicular to x -axis so $Q({x}_{1},-{y}_{1})$

It is given that OPQ is equilateral triangle where O is origin

so, $\anglePOQ={60}^{\circ }\implies \anglePOX={30}^{\circ }$

$	an {30}^{\circ }=$ slope of $OP=\frac{{y}_{1}}{{x}_{1}}$

$\frac{{y}_{1}}{{x}_{1}}=\frac{1}{\sqrt{3}}\implies {y}_{1}=\frac{{x}_{1}}{\sqrt{3}}$

Point $P({x}_{1},{y}_{1})$ satisfy hyperbola.

$\begin{gathered} \frac{{x}_{2}^{1}}{4}-\frac{{y}_{2}^{1}}{8}=1 \ \frac{{x}_{2}^{1}}{4}-\frac{{x}_{2}^{1}}{24}=1{{y}_{1}=\frac{{x}_{1}}{\sqrt{3}}} \ 6{x}_{2}^{1}-{x}_{2}^{1}=24 \ {x}_{2}^{1}=\frac{24}{5} \end{gathered}$

$\begin{gathered} 	riangleOPQ=\frac{1}{2}	imes PQ	imes ({x}_{1}) \ =\frac{1}{2}	imes 2{y}_{1}	imes {x}_{1}=\frac{{x}_{2}^{1}}{\sqrt{3}} \ =\frac{24}{5\sqrt{3}}	imes \frac{\sqrt{3}}{\sqrt{3}}=\frac{8\sqrt{3}}{5} \end{gathered}$

Solutions · 1