Assume that you wish to draw an ellipse with a semi-major axis of a = 35 cm and e = 0.25 What will the distance between the foci be?

Question

jamesj · Answer

Hello.  You should post this question in Mathematics, if you don't mind.

anonymous · Answer

This is actually my physics homework, but thanks, I might get better response there.

jamesj · Answer

Well, let's answer it here anyway.
For an ellipse
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} =  1 $$ (where we'll assume without loss of generality that a > b) the eccentricity is given by
$$ e = \sqrt{1 - b^2/a^2} $$
Now for that ellipse, what are the coordinates of the foci?

anonymous · Answer

I'm not sure what a and b represent? are those coordinates for the foci?

jamesj · Answer

No, that's the Cartesian form of an ellipse. Very important formula worth memorizing. Have a look here: http://en.wikipedia.org/wiki/Ellipse#Mathematical_definitions_and_properties the focal length f is $$ f = ae = \sqrt{a^2 - b^2} $$

anonymous · Answer

or a and b are the coordinates of the semimajor and minor axis?

anonymous · Answer

Thank you

jamesj · Answer

The parameters a and b are the lengths of the semi major and semi minor axes.  You're told here that a = 35 cm and e = 0.25.  Hence you can solve for b and thus for f.

anonymous · Answer

f=8.75
f is the distance of one focus to the center, so 2f=17.5, the distance between the foci?

jamesj · Answer

Yes, that sounds right.  I haven't run the calculations myself, but directionally, that's correct.