A) circle
B) an ellipse
C) a straight line
D) None of these
Correct Answer: C
Solution :
We have, \[|z+8|+|z-8|=16\] \[\Rightarrow \] \[|x+iy+8|+|x+iy-8|=16\] \[\Rightarrow \] \[|(x+8)+iy|=16-|(x-8)+iy|\] \[\Rightarrow \] \[\sqrt{{{(x+8)}^{2}}+{{y}^{2}}}=16\sqrt{{{(x-8)}^{2}}+{{y}^{2}}}\] On squaring both sides, we get \[{{(x+8)}^{2}}+{{y}^{2}}={{(16)}^{2}}+(\sqrt{{{(x-8)}^{2}}+{{y}^{2}}{{)}^{2}}}\] \[-2\times 16\times \sqrt{{{(x-8)}^{2}}+{{y}^{2}}}\] \[\Rightarrow \] \[{{x}^{2}}+64+16x+{{y}^{2}}=256+{{x}^{2}}+64\] \[-16x+{{y}^{2}}-32\sqrt{{{(x-8)}^{2}}+{{y}^{2}}}\] \[\Rightarrow \] \[32x=32[8-\sqrt{{{(x-8)}^{2}}+{{y}^{2}}}]\] \[\Rightarrow \] \[\sqrt{{{(x-8)}^{2}}+{{y}^{2}}}=8-x\] On squaring both sides, we get \[{{(x-8)}^{2}}{{y}^{2}}={{(8-x)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+64-16x+{{y}^{2}}=64+{{x}^{2}}-16x\] \[\Rightarrow \] \[{{y}^{2}}=0\Rightarrow \,\,\,y=0\]
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