Question

The eccentricity of the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ is

• A. $\frac{5}{3}$
• B. $\frac{4}{3}$
• C. $\frac{5}{4}$
• D. $\frac{4}{5}$

Answer 1

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

$\frac{x^2}{9}-\frac{y^2}{16}=1$
The eccentricity is given by,
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$