Question Number 43190 by MASANJA J last updated on 08/Sep/18
$${a}\:{point}\:{move}\:{in}\:{such}\:{away}\:{that}\:{its}\: \\ $$$${its}\:{distance}\:{from}\:{the}\:{x}−{axis}\:{is}\:{alwa} \\ $$$${yas}\frac{\mathrm{1}}{\mathrm{5}}\:{its}\:{distance}\:{from}\:{origin}. \\ $$$${find}\:{the}\:{equetion}\:{of}\:{its}\:{path}. \\ $$
Commented by MrW3 last updated on 08/Sep/18
$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{24}}}={k} \\ $$$${eqn}.: \\ $$$${y}=\pm{kx} \\ $$$${with}\:{k}=\frac{\mathrm{1}}{\sqrt{\mathrm{24}}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18
$${let}\:{the}\:{point}\:{be}\left(\alpha,\beta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }\:=\beta \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\mathrm{25}\beta^{\mathrm{2}} \\ $$$${so}\:{the}\:{locus}\:{is} \\ $$$$\propto^{\mathrm{2}} −\mathrm{24}\beta^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{24}{y}^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18
$${yes}... \\ $$