Question

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Eccentricity of the given hyperbola is

• A. $\frac{5}{3}$
• B. $\frac{3}{5}$
• C. $\frac{1}{3}$
• D. none of these

Answer 1

The correct option is A $\frac{5}{3}$

$16x^2-9y^2=144$

$\frac{x^2}{9}-\frac{y^2}{16}=1$

We know, $e=\sqrt{1+\frac{b^2}{a^2}}$

$e=\sqrt{1+\frac{16}{9}}$

$e=\sqrt{\frac{25}{9}}$

$e=\frac{5}{3}$

Hence, option A is correct.