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

Answer 1

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

If OPQ is equilateral triangle then OP makes ${30}^0$ with x-axis. $(\frac{\sqrt{3}r}{2},\frac{r}{2})\text{lies on hyperbola}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow r^2=\frac{16a^2{b}^2}{12b^2-4a^2}>0\Rightarrow 12b^2-4a^2>0\Rightarrow \frac{b^2}{a^2}>\frac{4}{12}{e}^2-1>\frac{1}{3}{e}^2>\frac{4}{3}\Rightarrow e>\frac{2}{\sqrt{3}}$