Question

If eccentricity of hyperbola and parabola are $e_1$ and $e_2$ respectively then $e_2–e_1$ equals

• A. $\frac{5-\sqrt{41}}{5}$
• B. $-\frac{8}{5}$
• C. $-\left(\frac{\sqrt{41}+5}{5}\right)$
• D. $\frac{\sqrt{41}-5}{5}$

Answer 1

The correct option is A $\frac{5-\sqrt{41}}{5}$

Eccentricity of hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ is
$\sqrt{1+\frac{16}{25}}=\frac{\sqrt{41}}{5}=e_1$
and eccentricity of parabola $=1=e_2$
$\therefore e_2-e_1=1-\frac{\sqrt{41}}{5}=\frac{5-\sqrt{41}}{5}$