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

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

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

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

TrifidBlue

and that's what I've said
thank you