Question

If AB is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$= 1 such that DOAB (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. e >$\frac{2}{\sqrt{3}}$

Answer 1

Answer: (D)

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

Answer 2

Let AB = 2l, then AM = 1

Thus, we have A = $\left\lbrack \frac{a\sqrt{b^{2} + l^{2}}}{b},l \right\rbrack$

Now, since OAB is an equilateral triangle,

Therefore we have, OA = 2l i.e. OM² + AM²

= (2l)² i.e. $\frac{a^{2}(b^{2} + l^{2})}{b^{2}}$ + l² = 4l²

Gives l² = $\frac{a^{2}b^{2}}{3b^{2} - a^{2}}$ > 0 i.e. 3b² – a² > 0 i.e. 3a² (e² – 1) > a² Gives e > $\frac{2}{\sqrt{3}}$.