Question

If the eccentricity of a hyperbola is 5/3, then the eccentricity of its conjugate is

• A. 44684
• B. 44685
• C. 5
• D. non existent

Answer 1

Answer: (B)

44685

Answer 2

Eccentricity of hyperbola is $\frac{5}{3}$
Let $e_1$ = eccentricity of hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $e_2$ = eccentricity of its conjugate $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$
$e_{1} = \sqrt{\frac{b^{2}}{a^{2}} +1}$ and $e_{2} = \sqrt{\frac{a^{2}}{b^{2}} + 1}$ given that $e_{1} = \frac{5}{3}$
$\therefore \:\:\: \frac{25}{9} =\frac{b^{2}}{a^{2}} + 1 \Rightarrow \frac{b^{2}}{a^{2}} + 1 \Rightarrow \frac{b^{2}}{a^{2}} = \frac{25}{9} - 1 = \frac{16}{9}$
$\therefore \:\:\: e_{2} = \sqrt{\frac{9}{16}+1} = \sqrt{\frac{25}{16} } = \frac{5}{4}$