Question

If PQ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$such that OPQ is an equilateral triangle, O being the centre of the hyperbola. Then, the eccentricity e of the hyperbola satisfies

• A. $1<e<2\sqrt{3}$
  
• B. $e=\frac{3}{\sqrt{3}}$
  
• C. $e=\frac{\sqrt{3}}{2}$
• D. $e=\frac{2}{\sqrt{3}}$

Answer 1

The correct option is D $e=\frac{2}{\sqrt{3}}$

$(OP{)}^2-(OQ{)}^2=(PQ{)}^2$
$\Rightarrow$ $\frac{a^2}{b^2}(b^2+l^2)+l^2=\frac{a^2}{b^2}(b^2+l^2)+l^2=4l^2$
$\Rightarrow$ $l^2=\frac{a^2{b}^2}{(3b^2-a^2)}>0$
$\Rightarrow$ $3b^2-a^2>0$
$\Rightarrow$ $3b^2>a^2\Rightarrow 3a^2(e^2-1)>a^2\Rightarrow e^2>\frac{4}{3}$
$\Rightarrow$ $e>\frac{2}{\sqrt{3}}$