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. zero

Answer 1

Answer: (B)

Key Idea: Eccentricity of hyperbola $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$ is $e=\sqrt{\frac{(a^2+b^2)}{a^2}}$ Given hyperbola $9x^2-16y^2-18x-64y-199=0$ $-16(y^2+4y+4-4)-199=0$ $\Rightarrow$ $9{(x-1)}^2-16{(y+2)}^2=144$ $\Rightarrow$ $\frac{{(x-1)}^2}{16}-\frac{{(y+2)}^2}{9}=1$ $\Rightarrow$ $a^2=16,b^2=9$ $\therefore$ $e=\sqrt{\frac{16+9}{16}}=\frac{5}{4}$ Note: Since eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ is same. $\therefore$ Eccentricity of hyperbola is independent of shift of centre.