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

Hint:

Use parametric coordinates and equilateral triangle property.

Answer 1

The Correct Option is D

Step 1: Formula / Definition}
\[ P(a\sec\theta, b\tan\theta), Q(a\sec\theta, -b\tan\theta), O(0,0) \]
Step 2: Calculation / Simplification}
\(OP = OQ = PQ\) for equilateral triangle
\(OP^2 = a^2\sec^2\theta + b^2\tan^2\theta\)
\(PQ = 2b\tan\theta\)
\(OP = PQ \Rightarrow a^2\sec^2\theta + b^2\tan^2\theta = 4b^2\tan^2\theta\)
\(a^2\sec^2\theta = 3b^2\tan^2\theta \Rightarrow \frac{a^2}{b^2} = 3\sin^2\theta\cos^2\theta\)
\(e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1}{3\sin^2\theta\cos^2\theta} = 1 + \frac{4}{3\sin^2 2\theta}\)
Since \(\sin^2 2\theta \leq 1\), \(e^2 \geq 1 + \frac{4}{3} = \frac{7}{3}>\frac{4}{3}\)
\(\therefore e>\frac{2}{\sqrt{3}}\)
Step 3: Final Answer
\[ e>\frac{2}{\sqrt{3}} \]