Question about semi-major axis of an ellipse

[TrifidBlue]

I hope this the right place to post my question...

should it be, "we can define a as half the sum of distances..."?
please correct and explain if I'm mistaken
thanks

 

[HallsofIvy]

You are mistaken.

When \theta= \theta', 1/r= C(1- \epsilon cos(\theta- \theta')=C(1- \epsilon cos(0))=C(1- \epsilon) so that r= 1/(C(1-\epsilon).

When \theta= \theta'+ \pi, 1/r= C(1+\epsilon cos(\theta- \theta- \pi)=C(1- \epsilon cos(\pi))=C(1+ \epsilon) so that r= 1/(C(1+\epsilon).

The total distance between those points is 1/C(1+\epsilon)+1/C(1- \epsilon)= (1/C)(1/(1+\epsilon)+ 1/(1- \epsilon)). Getting the common denominator, (1+\epsilon)(1-\epsilon)= 1-\epsilon^2, we have
(1/C)\frac{1- \epsilon+ 1+ \epsilon}{1- \epsilon^2}= (1/C)\frac{2}{1- \epsilon^2}

Half of that is
a= \frac{1}{C(1- \epsilon^2)}

 

[TrifidBlue]

You are mistaken.

When \theta= \theta', 1/r= C(1- \epsilon cos(\theta- \theta')=C(1- \epsilon cos(0))=C(1- \epsilon) so that r= 1/(C(1-\epsilon).

When \theta= \theta'+ \pi, 1/r= C(1+\epsilon cos(\theta- \theta- \pi)=C(1- \epsilon cos(\pi))=C(1+ \epsilon) so that r= 1/(C(1+\epsilon).

The total distance between those points is 1/C(1+\epsilon)+1/C(1- \epsilon)= (1/C)(1/(1+\epsilon)+ 1/(1- \epsilon)). Getting the common denominator, (1+\epsilon)(1-\epsilon)= 1-\epsilon^2, we have
(1/C)\frac{1- \epsilon+ 1+ \epsilon}{1- \epsilon^2}= (1/C)\frac{2}{1- \epsilon^2}

Half of that is
a= \frac{1}{C(1- \epsilon^2)}

and that's what I've said
thank you