Question

The average length of all vertical chords of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\),  \(a \leq x\leq 2a\), is

• A. b{\(2\sqrt3\)-ln(\(2+\sqrt3\))}
• B. b{\(3\sqrt2\)+ln(\(3+\sqrt2\))}
• C. b{\(2\sqrt5\)-ln(\(2+\sqrt5\))}
• D. b{\(5\sqrt2\)+ln(\(5+\sqrt2\))}

Answer 1

The Correct Option is A

Given:
Hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) 

Solving for y:
\[ y = \pm b\sqrt{\frac{x^2}{a^2} - 1} \]
Length of a vertical chord at x:
\[ \text{Chord length} = 2b\sqrt{\frac{x^2}{a^2} - 1} \]
Average chord length from \( x = a \) to \( x = 2a \):
\[ \text{Average length} = \frac{1}{a} \int_a^{2a} 2b\sqrt{\frac{x^2}{a^2} - 1} \, dx = \frac{2b}{a} \int_a^{2a} \sqrt{\frac{x^2}{a^2} - 1} \, dx \]
Substitute: \( x = a\sec\theta \), \( dx = a\sec\theta\tan\theta \, d\theta \)
Limits: \( x = a \Rightarrow \theta = 0 \), \( x = 2a \Rightarrow \theta = \sec^{-1}(2) \)
\[ \Rightarrow \int_a^{2a} \sqrt{\frac{x^2}{a^2} - 1} \, dx = \int_0^{\sec^{-1}(2)} \tan^2\theta \cdot a\sec\theta\tan\theta \, d\theta \quad \text{(not shown here; result used directly)} \]
Final result from solving the integral:
\[ \frac{2b}{a^2} \int_a^{2a} \sqrt{x^2 - a^2} \, dx = b(2\sqrt{3} - \ln(2 + \sqrt{3})) \]
Final Answer:
\[ \boxed{b(2\sqrt{3} - \ln(2 + \sqrt{3}))} \]
Correct option: (A)