Question Number 51271 by peter frank last updated on 25/Dec/18
$${Show}\:{that}\:{the}\:{locus}\:{of}\:{a} \\ $$$${point}\:{which}\:{moves}\:{so} \\ $$$${that}\:{its}\:{distance}\:{from} \\ $$$${the}\:{point}\:\left({ae},\mathrm{0}\right)\:{is}\:{e}\:{times} \\ $$$${its}\:{distance}\:{from}\:{the}\: \\ $$$${line}\:{x}=\frac{{a}}{{e}}\:{is}\:{given}\:{by}\:{the} \\ $$$${equation} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)}=\mathrm{1} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18
$$\sqrt{\left({x}−{ae}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }\:={e}\left({x}−\frac{{a}}{{e}}\right) \\ $$$$\sqrt{\left({x}−{ae}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\left({ex}−{a}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xae}+{a}^{\mathrm{2}} {e}^{\mathrm{2}} +{y}^{\mathrm{2}} ={e}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{xae}+{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right) \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)}=\mathrm{1} \\ $$$$ \\ $$
Commented by peter frank last updated on 25/Dec/18
$${much}\:{respect}\:{mr}\:{tanmay}. \\ $$$${GOD}\:{bless}\:{you}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18
$${thank}\:{you}... \\ $$