Question

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

• A. $e>\sqrt{3}$
• B. $1<e<\frac{2}{\sqrt{3}}$
• C. $e>\frac{2}{\sqrt{3}}$
• D. None of these

Answer 1

The correct option is C $e>\frac{2}{\sqrt{3}}$

Let the coordinates of A be $(\alpha ,\beta )$.
Then $AB=2\beta$ and $OA=\sqrt{{\alpha }^2+{\beta }^2}$
Since OAB is an equilateral triangle OA=AB
$\Rightarrow$ ${\alpha }^2+{\beta }^2=4{\beta }^2$ $\Rightarrow$ ${\alpha }^2=3{\beta }^2$
$\Rightarrow$ $\alpha =\pm \sqrt{3}\beta$
Also since $(\alpha ,\beta )$ lies on the given hyperbola,
$\frac{{\alpha }^2}{a^2-\frac{{\beta }^2}{b^2}}=1$
$\Rightarrow$ $\frac{3{\beta }^2}{a^2}-\frac{{\beta }^2}{b^2}=1$
$\Rightarrow \frac{3}{a^2}-\frac{1}{b^2}=\frac{1}{{\beta }^2}>0$
$\Rightarrow \frac{b^2}{a^2}>\frac{1}{3}$
$\Rightarrow$ $e^2-1>\frac{1}{3}$
$\Rightarrow e^2>\frac{4}{3}$
$\Rightarrow$ $e>\frac{2}{\sqrt{3}}$