Question

If the product of the perpendicular distances from any point on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) to its asymptotes is 6, and the eccentricity of the hyperbola is \( \sqrt{3} \), then the length of the conjugate axis is:

• A. \( 3 \)
• B. \( 6 \)
• C. \( 8 \)
• D. \( 12 \)

Hint:

Hyperbola Asymptotes and Conjugate Axis}
Product of perpendiculars to asymptotes is \( \frac{b^2}{a} \)
Use eccentricity relation: \( e^2 = 1 + \frac{b^2}{a^2} \)
Conjugate axis = \( 2b \), so find \( b^2 \) first using both identities

Answer 1

The Correct Option is B

Asymptotes of hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are: \[ \frac{x}{a} \pm \frac{y}{b} = 0 \] Product of perpendicular distances from any point on hyperbola to asymptotes is: \[ \text{Product} = \frac{b^2}{a} \quad \text{[standard result]} \Rightarrow \frac{b^2}{a} = 6 \tag{1} \] Also given: eccentricity \( e = \sqrt{3} \Rightarrow e^2 = 3 \Rightarrow \) \[ e^2 = 1 + \frac{b^2}{a^2} \Rightarrow 3 = 1 + \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = 2 \tag{2} \] From (1) and (2): \[ \frac{b^2}{a^2} = 2,\quad \frac{b^2}{a} = 6 \Rightarrow \frac{6}{a} = 2 \Rightarrow a = 3 \Rightarrow b^2 = 18 \Rightarrow b = \sqrt{18} \Rightarrow \text{Conjugate axis} = 2b = 2\sqrt{18} \approx 6 \Rightarrow \boxed{6} \]