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 triangle O being the origin, then the eccentricity of the hyperbola satisfies

• A. e >
• B. 1 < e <
• C.
• D. e >

Answer 1

The correct option is D

e >

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

$\therefore AB=2^{l}andAM=BM=l$

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

$OA=AB=OB=2^l$

$Also,O{M}^2+A{M}^2=O{A}^2\therefore \frac{a(b^2+l^2)}{b}+l^2=4l^2$

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

$Since{l}^2$ > 0 $\therefore \frac{a^2{b}^2}{3b^2-a^2}$ > $0\Rightarrow 3b^2-a^2$ >0

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