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TS EAMCET · Maths · Hyperbola

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 \(a-b=\)

  1. A \(4\)
  2. B \(3\)
  3. C \(2\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1\)

Step-by-step Solution

Detailed explanation

\(d_1 d_2 = \frac{a^2b^2}{a^2+b^2}\) \(\frac{a^2b^2}{a^2+b^2} = \frac{36}{13}\) \(e^2 = 1 + \frac{b^2}{a^2}\) \(\left(\frac{\sqrt{13}}{3}\right)^2 = 1 + \frac{b^2}{a^2}\) \(\frac{13}{9} = 1 + \frac{b^2}{a^2}\) \(\frac{b^2}{a^2} = \frac{4}{9} \implies b = \frac{2}{3}a\)…