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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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