- #1
Erez
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Hello,
given (x^2)/(a^2) + (y^2)/(b^2) = 1.
and using polar coordinates x=rcos(phi) , y=rsin(phi),
equating gives r^2 = 1/[(cos^2(phi)/a^2) + (sin^2(phi)/b^2)].
or if we leave b in the nominator :
r= b/[(sin^2(phi)+(b^2/a^2)cos^2(phi)]^1/2.
-could someone give a hint as to how the demoninator of the last expression can be turned into [1 - (e^2)cos^2(phi)]^1/2 where e is the eccentricity of the ellipse?
and what is the value of e?
thank you.
given (x^2)/(a^2) + (y^2)/(b^2) = 1.
and using polar coordinates x=rcos(phi) , y=rsin(phi),
equating gives r^2 = 1/[(cos^2(phi)/a^2) + (sin^2(phi)/b^2)].
or if we leave b in the nominator :
r= b/[(sin^2(phi)+(b^2/a^2)cos^2(phi)]^1/2.
-could someone give a hint as to how the demoninator of the last expression can be turned into [1 - (e^2)cos^2(phi)]^1/2 where e is the eccentricity of the ellipse?
and what is the value of e?
thank you.
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