Question

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

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

Answer 1

Answer: (B)

Eccentricity of hyperbola
$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$
is $e=\sqrt{\frac{\left(a^2+b^2\right)}{a^2}}$
Given hyperbola
$9x^2-16y^2-18x-64y-199=0$
$\Rightarrow -9\left(x^2-2x+1-1\right)$
$-16\left(y^2+4y+4-4\right)-199=0$
$\Rightarrow 9(x-1{)}^2-16(y+2{)}^2=144$
$\Rightarrow \frac{(x-1{)}^2}{16}-\frac{(y+2{)}^2}{9}=1$
$\Rightarrow a^2=16,b^2=9$
$\therefore e=\sqrt{\frac{16+9}{16}}=\frac{5}{4}$