Question

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

• A. $\frac{16}{9}$
• B. $\frac{5}{4}$
• C. $25 \quad$
• 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
$9 x^{2}-16 y^{2}-18 x-64 y-199=0$
$\Rightarrow \,-9\left(x^{2}-2 x+1-1\right)$
$-16\left(y^{2}+4 y+4-4\right)-199=0$
$\Rightarrow \, 9(x-1)^{2}-16(y+2)^{2}=144$
$\Rightarrow \, \frac{(x-1)^{2}}{16}-\frac{(y+2)^{2}}{9}=1$
$\Rightarrow \,a^{2}=16, b^{2}=9$
$\therefore \, e=\sqrt{\frac{16+9}{16}}=\frac{5}{4}$