Question Number 47438 by rahul 19 last updated on 10/Nov/18
$${Find}\:{locus}\:{of}\:{a}\:{point}\:{P}\:{which}\:{moves} \\ $$$${such}\:{that}\:{its}\:{distance}\:{from}\:{the}\:{line} \\ $$$${y}=\sqrt{\mathrm{3}}{x}−\mathrm{7}\:{is}\:{same}\:{as}\:{its}\:{distance}\:{from} \\ $$$$\left(\mathrm{2}\sqrt{\mathrm{3}},−\mathrm{1}\right)\:? \\ $$
Answered by ajfour last updated on 10/Nov/18
$$\frac{\left({y}−{x}\sqrt{\mathrm{3}}+\mathrm{7}\right)^{\mathrm{2}} }{\mathrm{4}}=\left({x}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{49}−\mathrm{2}\sqrt{\mathrm{3}}{xy}−\mathrm{14}\sqrt{\mathrm{3}}{x} \\ $$$$\:+\mathrm{14}{y}\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{16}\sqrt{\mathrm{3}}{x}+\mathrm{8}{y}+\mathrm{52} \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{xy}−\mathrm{2}\sqrt{\mathrm{3}}{x}−\mathrm{6}{y}+\mathrm{3}\:=\mathrm{0}\: \\ $$$$\Rightarrow\:\left({x}+\sqrt{\mathrm{3}}{y}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\left({x}+\sqrt{\mathrm{3}}{y}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left({x}+\sqrt{\mathrm{3}}{y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\sqrt{\mathrm{3}}{y}\:=\:\sqrt{\mathrm{3}}−{x}\:. \\ $$
Commented by rahul 19 last updated on 10/Nov/18
thanks sir����
Answered by rahul 19 last updated on 10/Nov/18
$${since}\:{the}\:{point}\:{satisifes}\:{the}\:{line}. \\ $$$${locus}\:{is}\:{st}.\:{line}\:{perpendicular}\:{to}\:{given} \\ $$$${line}\:{passing}\:{through}\:{given}\:{point}. \\ $$$$\Rightarrow\:{locus}\:=\:\left({y}+\mathrm{1}\right)=\:−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left({x}−\mathrm{2}\sqrt{\mathrm{3}}\right).\: \\ $$