Question

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

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

Answer 1

The correct option is 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
$\frac{x^2}{a^2}-\frac{l^2}{b^2}=1\Rightarrow x=\frac{a\sqrt{b^2+l^2}}{b}$
$\therefore Ais(\frac{a\sqrt{b^2+l^2}}{b},l)$
Since $\Delta OAB$ is equilateral triangle, therefore
$OA+AB+OB+2^l$
Also, $O{M}^2+A{M}^2=O{A}^2\therefore \frac{a(b^2+l^2)}{b}+l^2=4l^{2}Weget{l}^2=\frac{a^2{b}^2}{3b^2-a^2}Since{l}^2>O\therefore \frac{a^2{b}^2}{3b^2-a^2}>0\Rightarrow 3b^2-a^2>0\Rightarrow 3a^2(e^2-1)-a^2>0\Rightarrow e>\frac{2}{\sqrt{3}}$