AP EAMCET · Maths · Hyperbola
If the distance between the foci of a hyperbola \(H\) is 26 and distance between its directrices is \(\frac{50}{13}\) then the eccentricity of the conjugate hyperbola of the hyperbola \(H\) is
- A \(\frac{13}{12}\)
- B \(\frac{25}{17}\)
- C \(\frac{13}{7}\)
- D \(\frac{25}{13}\)
Answer & Solution
Correct Answer
(A) \(\frac{13}{12}\)
Step-by-step Solution
Detailed explanation
\(ae = \frac{26}{2} = 13\) \(\frac{a}{e} = \frac{1}{2} \cdot \frac{50}{13} = \frac{25}{13}\) \(a^2 = (ae) \cdot \left(\frac{a}{e}\right) = 13 \cdot \frac{25}{13} = 25\) \(e = \frac{13}{a} = \frac{13}{5}\) For conjugate hyperbola eccentricity \(e'\):…
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