A) 1
B) \[\sqrt{2}\]
C) 2
D) \[1/2\]
Correct Answer: B
Solution :
We have,\[{{x}^{2}}-{{y}^{2}}-4x+4y+16=0\] \[\Rightarrow \] \[({{x}^{2}}-4x)-({{y}^{2}}-4y)=-16\] \[\Rightarrow \]\[({{x}^{2}}-4x+4)-({{y}^{2}}-4y+4)=-16\] \[\Rightarrow \] \[{{(x-2)}^{2}}-{{(y-2)}^{2}}=-16\] \[\Rightarrow \] \[\frac{{{(x-2)}^{2}}}{{{4}^{2}}}-\frac{{{(y-2)}^{2}}}{{{4}^{2}}}=1\] Shifting the origin at (2, 2), we get \[\frac{{{X}^{2}}}{{{4}^{2}}}-\frac{{{Y}^{2}}}{{{4}^{2}}}=-1\]where \[x=X+2,\text{ }y=Y+2\] This is a rectangular hyperbola, whose eccentricity is always\[\sqrt{2}\].
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