Question about semi-major axis of an ellipse

[TrifidBlue]

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

http://i912.photobucket.com/albums/ac325/marwaibrahim/2.jpg

http://i912.photobucket.com/albums/ac325/marwaibrahim/1-2.jpg?t=1278189594

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 :confused:
thank you