March | 2015 | aussiemathstutor

$\begin{array}{l}\left\{ {z:\left| {z+1-i} \right|+\left| {z+1-i} \right|=4} \right\}\\\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}}}+\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}=4\\\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}}}=4-\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\{{\left( {\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y-1} \right)}}^{2}}}}} \right)}^{2}}={{\left( {4-\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}} \right)}^{2}}\\{{\left( {x+1} \right)}^{2}}+{{\left( {y-1} \right)}^{2}}=16-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}+{{\left( {x+1} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}\\{{\left( {y-1} \right)}^{2}}-{{\left( {y+1} \right)}^{2}}-16=-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\{{y}^{2}}-2y+1-\left( {{{y}^{2}}+2y+1} \right)-16=-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\-4y-16=-8\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}\\{{\left( {-y-4} \right)}^{2}}={{\left( {-2\sqrt{{{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}}}} \right)}^{2}}\\{{y}^{2}}+8y+16=4\left( {{{{\left( {x+1} \right)}}^{2}}+{{{\left( {y+1} \right)}}^{2}}} \right)\\=4{{\left( {x+1} \right)}^{2}}+4{{y}^{2}}+8y+4\\\frac{{12=4{{{\left( {x+1} \right)}}^{2}}+4{{y}^{2}}}}{{12}}\\\frac{{{{{\left( {x+1} \right)}}^{2}}}}{3}+{{y}^{2}}=1\end{array}$

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