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 \( \frac{36}{13} \) and its eccentricity is \( \frac{\sqrt{13}}{3} \), then \( a - b = \)

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Always simplify the eccentricity relation first to express \( b \) in terms of \( a \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

We use the standard property of hyperbolas regarding the product of perpendiculars to asymptotes and the formula connecting eccentricity, \( a \), and \( b \).
Step 2: Key Formula or Approach:

1. Product of perpendiculars \( P = \frac{a^2 b^2}{a^2 + b^2} \). 2. Eccentricity relation \( b^2 = a^2(e^2 - 1) \).
Step 3: Detailed Explanation:

Given \( e = \frac{\sqrt{13}}{3} \). Relation: \( b^2 = a^2 \left( \frac{13}{9} - 1 \right) = a^2 \left( \frac{4}{9} \right) \). So \( b = \frac{2}{3}a \). Given product \( P = \frac{36}{13} \). Substitute \( b^2 = \frac{4}{9}a^2 \) into the product formula: \[ \frac{a^2 (\frac{4}{9}a^2)}{a^2 + \frac{4}{9}a^2} = \frac{36}{13} \] \[ \frac{\frac{4}{9}a^4}{\frac{13}{9}a^2} = \frac{36}{13} \] \[ \frac{4a^2}{13} = \frac{36}{13} \] \[ 4a^2 = 36 \implies a^2 = 9 \implies a = 3 \] Calculate \( b \): \[ b = \frac{2}{3}(3) = 2 \] Required value: \( a - b = 3 - 2 = 1 \).
Step 4: Final Answer:

The value of \( a-b \) is 1.