Question

The eccentricity of the hyperbola $ 9{{x}^{2}}-16{{y}^{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 $ 9{{x}^{2}}-16{{y}^{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.