Question

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. \( \frac{25}{13} \)

Hint:

For hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \): Distance between foci = \(2ae\). Distance between directrices = \(2a/e\). Eccentricity \(e\). Relation \(b^2 = a^2(e^2-1)\). For its conjugate hyperbola \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), let eccentricity be \(e'\). Then \(a^2 = b^2((e')^2-1)\). A useful relation: \( \frac{1}{e^2} + \frac{1}{(e')^2} = 1 \).

Answer 1

The Correct Option is A

Let the hyperbola H be \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Let its eccentricity be \(e\).
Distance between foci = \( 2ae \).
Given \( 2ae = 26 \implies ae = 13 \cdots (1) \).
Distance between directrices = \( \frac{2a}{e} \).
Given \( \frac{2a}{e} = \frac{50}{13} \implies \frac{a}{e} = \frac{25}{13} \cdots (2) \).
Multiply (1) and (2): \( (ae) \left(\frac{a}{e}\right) = 13 \times \frac{25}{13} \) \[ a^2 = 25 \implies a=5 \] Substitute \(a=5\) into (1): \( 5e = 13 \implies e = \frac{13}{5} \).
For a hyperbola, \( e>1 \).
\( 13/5 = 2.
6>1 \), so this is valid.
Now find \(b^2\) using \( b^2 = a^2(e^2-1) \).
\[ b^2 = 25 \left( \left(\frac{13}{5}\right)^2 - 1 \right) = 25 \left( \frac{169}{25} - 1 \right) = 25 \left( \frac{169-25}{25} \right) = 25 \left( \frac{144}{25} \right) = 144 \] So \(b=12\).
The hyperbola H is \( \frac{x^2}{25} - \frac{y^2}{144} = 1 \).
Let \( e' \) be the eccentricity of the conjugate hyperbola.
The conjugate hyperbola is \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).
Its eccentricity \( e' \) satisfies \( a^2 = b^2((e')^2-1) \) (roles of a and b are swapped for the formula \(b^2=a^2(e^2-1)\)).
So, \( (e')^2 = 1 + \frac{a^2}{b^2} \).
\[ (e')^2 = 1 + \frac{25}{144} = \frac{144+25}{144} = \frac{169}{144} \] \[ e' = \sqrt{\frac{169}{144}} = \frac{13}{12} \] Alternatively, the relation between eccentricities of a hyperbola and its conjugate is \( \frac{1}{e^2} + \frac{1}{(e')^2} = 1 \).
\[ \frac{1}{(13/5)^2} + \frac{1}{(e')^2} = 1 \] \[ \frac{25}{169} + \frac{1}{(e')^2} = 1 \] \[ \frac{1}{(e')^2} = 1 - \frac{25}{169} = \frac{169-25}{169} = \frac{144}{169} \] \[ (e')^2 = \frac{169}{144} \implies e' = \frac{13}{12} \] This matches option (1).