Question

The eccentricity of the hyperbola $9x^2-16y^2-18x-64y-199=0$ is

• A. $\frac{16}{9}$
• B. $\frac{5}{4}$
• C. $\frac{25}{16}$
• D. $0$

Answer 1

The correct option is B

$\frac{5}{4}$

Explanation for the correct option:

Given equation of the hyperbola,

$$\begin{gathered} \Rightarrow 9x^2-16y^2-18x-64y-199=0 \\ \Rightarrow 9\left[{\left(x-1\right)}^2-1\right]-16\left[{\left(y+2\right)}^2-4\right]=199 \\ \Rightarrow 9{\left(x-1\right)}^2-9-16{\left(y+2\right)}^2+64=199 \\ \Rightarrow 9{\left(x-1\right)}^2-16{\left(y+2\right)}^2=144 \\ \Rightarrow \frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+2\right)}^2}{9}=1 \end{gathered}$$

so,

$$\begin{gathered} a=4,b=3 \\ e=\sqrt{1+\frac{b^2}{a^2}} \\ =\sqrt{1+\frac{9}{16}} \\ =\sqrt{\frac{25}{16}} \\ =\frac{5}{4} \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer