Question Number 48143 by rahul 19 last updated on 20/Nov/18
$${The}\:{locus}\:{of}\:{P}\left({x},{y}\right)\:{such}\:{that} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{8}{y}+\mathrm{16}}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}}=\mathrm{5}\:{is}? \\ $$
Commented by rahul 19 last updated on 20/Nov/18
$${I}'{m}\:{getting}\:{parabola}\:{but}\:{ans}.\:{is}\:{infinite} \\ $$$${ray}. \\ $$
Answered by MJS last updated on 20/Nov/18
$$\sqrt{{A}}−\sqrt{{B}}=\mathrm{5} \\ $$$${A}+{B}−\mathrm{2}\sqrt{{A}}\sqrt{{B}}=\mathrm{25} \\ $$$$\mathrm{2}\sqrt{{A}}\sqrt{{B}}={A}+{B}−\mathrm{25} \\ $$$$\mathrm{4}{AB}={A}^{\mathrm{2}} +\mathrm{2}{AB}+{B}^{\mathrm{2}} −\mathrm{50}{A}−\mathrm{50}{B}+\mathrm{625} \\ $$$${A}^{\mathrm{2}} −\mathrm{2}{AB}+{B}^{\mathrm{2}} −\mathrm{50}{A}−\mathrm{50}{B}+\mathrm{625}=\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} −\mathrm{24}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{96}{x}+\mathrm{72}{y}+\mathrm{144}=\mathrm{0} \\ $$$$\left(\mathrm{4}{x}−\mathrm{3}{y}−\mathrm{12}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{y}=\frac{\mathrm{4}}{\mathrm{3}}{x}−\mathrm{4} \\ $$