Question

If $ e_1 $ and $ e_2 $ are the eccentricities of the hyperbola $$ 16x^2 - 9y^2 = 1 $$ and its conjugate respectively, then find $ 3e_1 $ in terms of $ e_2 $.

• A. \( 5e_2 \)
• B. \( 4e_2 \)
• C. \( 2e_2 \)
• D. \( e_2^2 \)

Hint:

Eccentricity of a hyperbola is \( e = \sqrt{1 + \frac{b^2}{a^2}} \), and for its conjugate it is \( \sqrt{1 + \frac{a^2}{b^2}} \).

Answer 1

The Correct Option is B

Given: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \Rightarrow \frac{x^2}{1/16} - \frac{y^2}{1/9} = 1 \Rightarrow a^2 = \frac{1}{16},\ b^2 = \frac{1}{9} \] For hyperbola: \[ e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1/9}{1/16} = 1 + \frac{16}{9} = \frac{25}{9} \Rightarrow e_1 = \frac{5}{3} \] Conjugate hyperbola: \[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \Rightarrow e_2^2 = 1 + \frac{a^2}{b^2} = 1 + \frac{1/16}{1/9} = 1 + \frac{9}{16} = \frac{25}{16} \Rightarrow e_2 = \frac{5}{4} \] Now: \[ 3e_1 = 3 \cdot \frac{5}{3} = 5,\quad 4e_2 = 4 \cdot \frac{5}{4} = 5 \Rightarrow \boxed{3e_1 = 4e_2} \]