Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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 (x^2 /A^2 )−(y^2 /B^2 )=1 B^2 =A^2 (e_2 ^2 −1) for ellipse b^2 =a^2 (1−e_1 ^2 ) foci of hyerbola =(Ae_2 ,0)and(−Ae_2 ,0) vertices of ellipse =(a,0) and(−a,0) foci of ellipse(ae_1 ,0) and(−ae_1 ,0) given ae_1 =A so (a/A)=(1/e_1 ) Ae_2 =a (a/A)=e_2 (1/e_1 )=e_2 ((b/B))^2 =((a^2 (1−e_1 ^2 ))/(A^2 (e_2 ^2 −1)))=((e_2 ^2 (1−(1/e_2 ^2 )))/((e_2 ^2 −1)))=1

$${for}\:{hyperbola}\:\:\frac{{x}^{\mathrm{2}} }{{A}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{B}^{\mathrm{2}} }=\mathrm{1}\:\:\:{B}^{\mathrm{2}} ={A}^{\mathrm{2}} \left({e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${for}\:{ellipse}\:\:{b}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}−{e}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$${foci}\:{of}\:{hyerbola}\:=\left({Ae}_{\mathrm{2}} ,\mathrm{0}\right){and}\left(−{Ae}_{\mathrm{2}} ,\mathrm{0}\right) \\ $$$${vertices}\:{of}\:{ellipse}\:=\left({a},\mathrm{0}\right)\:{and}\left(−{a},\mathrm{0}\right) \\ $$$${foci}\:{of}\:{ellipse}\left({ae}_{\mathrm{1}} ,\mathrm{0}\right)\:{and}\left(−{ae}_{\mathrm{1}} ,\mathrm{0}\right) \\ $$$${given}\:{ae}_{\mathrm{1}} ={A}\:\:\:\:{so}\:\frac{{a}}{{A}}=\frac{\mathrm{1}}{{e}_{\mathrm{1}} } \\ $$$${Ae}_{\mathrm{2}} ={a}\:\:\:\:\frac{{a}}{{A}}={e}_{\mathrm{2}} \:\: \\ $$$$\frac{\mathrm{1}}{{e}_{\mathrm{1}} }={e}_{\mathrm{2}} \\ $$$$\left(\frac{{b}}{{B}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}_{\mathrm{1}} ^{\mathrm{2}} \right)}{{A}^{\mathrm{2}} \left({e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}\right)}=\frac{{e}_{\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{e}_{\mathrm{2}} ^{\mathrm{2}} }\right)}{\left({e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}\right)}=\mathrm{1} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

e_1 +e_2 =((√e_1 ) −(√e_2 ) )^2 +2(√(e_1 e_2 )) =((√e_1 ) −(√e_2 ) )^2 +2 (since e_1 e_2 =1 already proved) e_1 +e_2 >2 so e_1 +e_2 can not be equals to 2

$${e}_{\mathrm{1}} +{e}_{\mathrm{2}} =\left(\sqrt{{e}_{\mathrm{1}} }\:−\sqrt{{e}_{\mathrm{2}} }\:\right)^{\mathrm{2}} +\mathrm{2}\sqrt{{e}_{\mathrm{1}} {e}_{\mathrm{2}} } \\ $$$$\:\:\:=\left(\sqrt{{e}_{\mathrm{1}} }\:−\sqrt{{e}_{\mathrm{2}} }\:\right)^{\mathrm{2}} +\mathrm{2}\:\:\left({since}\:{e}_{\mathrm{1}} {e}_{\mathrm{2}} =\mathrm{1}\:{already}\:{proved}\right) \\ $$$${e}_{\mathrm{1}} +{e}_{\mathrm{2}} >\mathrm{2}\:\:{so}\:{e}_{\mathrm{1}} +{e}_{\mathrm{2}} \:\:{can}\:{not}\:{be}\:{equals}\:{to}\:\mathrm{2} \\ $$

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 48 still interested in physics and math...look back...the year 1988. when i was in class XII..

$${its}\:{ok}...{at}\:{the}\:{age}\:{of}\:\mathrm{48}\:{still}\:{interested}\:{in} \\ $$$${physics}\:{and}\:{math}...{look}\:{back}...{the}\:{year}\:\mathrm{1988}. \\ $$$${when}\:{i}\:{was}\:{in}\:{class}\:{XII}.. \\ $$

Commented by rahul 19 last updated on 15/Jul/18

That's the spirit! ������������

Terms of Service

Privacy Policy

Contact: info@tinkutara.com