A) \[\frac{16}{9}\]
B) \[\frac{5}{4}\]
C) \[\frac{25}{16}\]
D) zero
Correct Answer: B
Solution :
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.
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