Question

If $AB$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $△OAB$ is an equilateral triangle, $O$ being the origin, then the eccentricity of the hyperbola satisfies :

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

Answer 1

Answer: (D)

$e>\frac{2}{\sqrt{3}}$

Let the length of the double ordinate be $2l$

$\therefore AB=2l$ and $AM=BM=l$

Clearly ordinate of point $A$ is $l$.

The abscissa of the point $A$ is given by

$\frac{x^2}{a^2}-\frac{l^2}{b^2}=1\Rightarrow x=\frac{a\sqrt{b^2+l^2}}{b}$

$\therefore A$ is $(\frac{a\sqrt{b^2+l^2}}{b},l)$

Since $△OAB$ is equilateral triangle, therefore

$OA=AB=OB=2l$

Also, ${OM}^2+{AM}^2={OA}^2\Rightarrow \frac{a^2(b^2+l^2)}{b^2}+l^2=4l^2$

$\therefore l^2=\frac{a^2{b}^2}{3b^2-a^2}$

Since $l^2>0$

$\therefore \frac{a^2{b}^2}{3b^2-a^2}>0\Rightarrow 3b^2-a^2>0\Rightarrow 3a^2(e^2-1)-a^2>0\Rightarrow e>\frac{2}{\sqrt{3}}$