Question

The eccentricity of the hyperbola $\frac{x^2}{16}-\frac{y^2}{25}=1$ is

• A. $\frac{3}{4}$
• B. $\frac{3}{5}$
• C. $\frac{\sqrt{41}}{4}$
• D. $\frac{\sqrt{41}}{5}$

Answer 1

The correct option is C

$\frac{\sqrt{41}}{4}$

Explanation for the correct option:

Finding the eccentricity:

Equation of the hyperbola is $\frac{x^2}{16}-\frac{y^2}{25}=1$.

General equation of hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$.

Comparing these two equations, we get

$a^2=16$ and $b^2=25$.

Then the eccentricity is calculated.

$$\begin{gathered} e=\sqrt{1+\frac{b^2}{a^2}} \\ =\sqrt{1+\frac{25}{16}} \\ =\sqrt{\frac{41}{16}} \\ =\frac{\sqrt{41}}{4} \end{gathered}$$

Therefore, the correct answer is option (C).