Question Number 39955 by rahul 19 last updated on 13/Jul/18
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18
$${for}\:{hyperbola}\:\:\frac{{x}^{\mathrm{2}} }{{A}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{B}^{\mathrm{2}} }=\mathrm{1}\:\:\:{B}^{\mathrm{2}} ={A}^{\mathrm{2}} \left({e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${for}\:{ellipse}\:\:{b}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}−{e}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$${foci}\:{of}\:{hyerbola}\:=\left({Ae}_{\mathrm{2}} ,\mathrm{0}\right){and}\left(−{Ae}_{\mathrm{2}} ,\mathrm{0}\right) \\ $$$${vertices}\:{of}\:{ellipse}\:=\left({a},\mathrm{0}\right)\:{and}\left(−{a},\mathrm{0}\right) \\ $$$${foci}\:{of}\:{ellipse}\left({ae}_{\mathrm{1}} ,\mathrm{0}\right)\:{and}\left(−{ae}_{\mathrm{1}} ,\mathrm{0}\right) \\ $$$${given}\:{ae}_{\mathrm{1}} ={A}\:\:\:\:{so}\:\frac{{a}}{{A}}=\frac{\mathrm{1}}{{e}_{\mathrm{1}} } \\ $$$${Ae}_{\mathrm{2}} ={a}\:\:\:\:\frac{{a}}{{A}}={e}_{\mathrm{2}} \:\: \\ $$$$\frac{\mathrm{1}}{{e}_{\mathrm{1}} }={e}_{\mathrm{2}} \\ $$$$\left(\frac{{b}}{{B}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}_{\mathrm{1}} ^{\mathrm{2}} \right)}{{A}^{\mathrm{2}} \left({e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}\right)}=\frac{{e}_{\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{e}_{\mathrm{2}} ^{\mathrm{2}} }\right)}{\left({e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}\right)}=\mathrm{1} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18
$${e}_{\mathrm{1}} +{e}_{\mathrm{2}} =\left(\sqrt{{e}_{\mathrm{1}} }\:−\sqrt{{e}_{\mathrm{2}} }\:\right)^{\mathrm{2}} +\mathrm{2}\sqrt{{e}_{\mathrm{1}} {e}_{\mathrm{2}} } \\ $$$$\:\:\:=\left(\sqrt{{e}_{\mathrm{1}} }\:−\sqrt{{e}_{\mathrm{2}} }\:\right)^{\mathrm{2}} +\mathrm{2}\:\:\left({since}\:{e}_{\mathrm{1}} {e}_{\mathrm{2}} =\mathrm{1}\:{already}\:{proved}\right) \\ $$$${e}_{\mathrm{1}} +{e}_{\mathrm{2}} >\mathrm{2}\:\:{so}\:{e}_{\mathrm{1}} +{e}_{\mathrm{2}} \:\:{can}\:{not}\:{be}\:{equals}\:{to}\:\mathrm{2} \\ $$
Commented by rahul 19 last updated on 15/Jul/18
Thank You Sir ! ������
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18
$${its}\:{ok}…{at}\:{the}\:{age}\:{of}\:\mathrm{48}\:{still}\:{interested}\:{in} \\ $$$${physics}\:{and}\:{math}…{look}\:{back}…{the}\:{year}\:\mathrm{1988}. \\ $$$${when}\:{i}\:{was}\:{in}\:{class}\:{XII}.. \\ $$
Commented by rahul 19 last updated on 15/Jul/18
That's the spirit! ������������