A) \[k\in (0,2)\]
B) \[k\in (0,1)\]
C) \[k\in (1,\infty )\]
D) \[k\in R'\]
Correct Answer: A
Solution :
Given that, \[\left| \sqrt{{{x}^{2}}+{{(y-1)}^{2}}-\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}} \right|=k\] it can be rewritten as \[\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}=k+\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}\] On squaring both sides, we get \[{{x}^{2}}+{{y}^{2}}-2y+1={{k}^{2}}+({{x}^{2}}+{{y}^{2}}+2y+1)\] \[+2k\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}\] \[\Rightarrow \]\[-4y-{{k}^{2}}=2k\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}\] Again squaring both sides, we get \[16{{y}^{2}}+{{k}^{4}}+8y{{k}^{2}}=4{{k}^{2}}\,\,({{x}^{2}}+{{y}^{2}}+1)\] \[\Rightarrow \] \[4{{x}^{2}}{{k}^{2}}+(4{{k}^{2}}-16){{y}^{2}}={{k}^{4}}-4{{k}^{2}}\] To represent an equation of hyperbola, the coefficient of either \[{{x}^{2}}\]or \[{{y}^{2}}\] is negative. But coefficient of \[{{x}^{2}}\] cannot be negative, so we take the coefficient of \[{{y}^{2}}\]. \[\therefore \] \[4{{k}^{2}}-16<0\] \[\Rightarrow \] \[{{k}^{2}}\le 4\Rightarrow -2<k<2\] As the given equation k cannot be negative. \[\therefore \] \[0<k<2\]
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