A) < 1
B) < 2
C) > 4
D) None of these
Correct Answer: C
Solution :
Let \[z=x+iy\] \[\therefore \]\[\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}+\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}=k\] ?(i) \[\Rightarrow \]\[{{x}^{2}}+{{(y-1)}^{2}}-{{x}^{2}}-{{(y+1)}^{2}}\] \[=k\{\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}-\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}\}\] \[\therefore \]\[\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}-\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}\] \[=-\frac{4y}{k}\] ?(ii) From Eqs. (i) and (ii), \[2\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}=k-\frac{4y}{k}\] \[\Rightarrow \]\[4{{x}^{2}}+\left( 4-\frac{16}{{{k}^{2}}} \right){{y}^{2}}={{k}^{2}}-4\] For an ellipse, \[4-\frac{16}{{{k}^{2}}}>0\Rightarrow {{k}^{2}}-4>0\] \[\Rightarrow \] \[{{k}^{2}}>0\]
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