A) \[-2<k<2\]
B) \[k>2\]
C) \[0<k<2\]
D) None of these
Correct Answer: A
Solution :
Let\[z=x+iy\], then\[|z+i|-|z-i|\,\,=k\] becomes \[\therefore \] \[\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}-\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}=k\]... (i) or \[{{x}^{2}}+{{(y+1)}^{2}}-{{x}^{2}}-{{(y-1)}^{2}}\] \[=k\{\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}+\sqrt{{{x}^{2}}{{(y-1)}^{2}}\}}\] \[\therefore \] \[\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}+\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}=\frac{4y}{k}\]... (ii) From Eqs. (i) and (ii), \[2\sqrt{{{x}^{2}}+{{(y+1)}^{2}}}=k+\frac{4y}{k}\] \[\Rightarrow \]\[4{{x}^{2}}+4{{y}^{2}}+8y+4={{k}^{2}}+\frac{16{{y}^{2}}}{{{k}^{2}}}+8y\] \[\Rightarrow \] \[4{{z}^{2}}+\left( 4-\frac{16}{{{k}^{2}}} \right){{y}^{2}}={{k}^{2}}-4\] For an hyperbola,\[\frac{4{{k}^{2}}-16}{{{k}^{2}}}<0\Rightarrow {{k}^{2}}-4<0\] \[\Rightarrow \] \[|k|\,\,<2\]or\[-2<k<2\]
You need to login to perform this action.
You will be redirected in
3 sec
