Question about semi-major axis of an ellipse

  • Thread starter TrifidBlue
  • Start date
  • #1
16
0
I hope this the right place to post my question...

2.jpg


1-2.jpg


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

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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]
 
Last edited by a moderator:
  • #3
16
0
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]
and that's what I've said :confused:
thank you
 

Related Threads on Question about semi-major axis of an ellipse

  • Last Post
Replies
2
Views
3K
Replies
1
Views
663
  • Last Post
Replies
2
Views
807
  • Last Post
Replies
11
Views
2K
Replies
5
Views
2K
Replies
27
Views
4K
Replies
4
Views
1K
Replies
3
Views
3K
Replies
6
Views
12K
  • Last Post
Replies
4
Views
7K
Top