Question Number 45451 by Tinkutara last updated on 13/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18
$$\mid\left({x}+{iy}−\mathrm{3}−\mathrm{2}{i}\right)\mid=\mid\left({x}+{iy}\right){cos}\left(\frac{\pi}{\mathrm{4}}−{tan}^{−\mathrm{1}} \frac{{y}}{{x}}\right)\mid \\ $$$$\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }\:=\mid\left({x}+{iy}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\right)\mid \\ $$$${calculation}\:{right}\:{hand}\:{side} \\ $$$${let}\:{k}=\left(\frac{{x}+{y}}{\:\sqrt{\mathrm{2}}\:×\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\right) \\ $$$$\mid{kx}+{iky}\mid \\ $$$$\sqrt{{k}^{\mathrm{2}} {x}^{\mathrm{2}} +{k}^{\mathrm{2}} {y}^{\mathrm{2}} }\left[\right. \\ $$$$={k}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\: \\ $$$$=\frac{{x}+{y}}{\:\sqrt{\mathrm{2}}×\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}×\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\: \\ $$$$=\frac{{x}+{y}}{\:\sqrt{}\mathrm{2}} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\frac{\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${pls}\:{wait}… \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}+{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{4}\right)−\left({x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{12}{x}−\mathrm{8}{y}+\mathrm{8}=\mathrm{0} \\ $$$${A}=\mathrm{1}\:\:\:{B}=−\mathrm{2}\:\:\:\:{C}=\mathrm{1}\:\:\:{D}=−\mathrm{1}\:\:\:{E}=−\mathrm{8}\:\:{F}=\mathrm{8} \\ $$$${B}^{\mathrm{2}} −\mathrm{4}{AC}=\mathrm{0} \\ $$$${so}\:{locus}\:{is}\:{parabola}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18
Commented by Tinkutara last updated on 13/Oct/18
Thank you very much Sir! I got the answer. ��������