Given foci and the ellipse's constant, how can I find the major and minor axes?

Question

Given that the definition of an ellipse is the set of points (x,y) where the distance from focus 1 to (x,y) plus the distance from from focus 2 to (x,y) is a constant, and given that constant, and the positions of the foci, how can I find the lengths of the major and minor axes of the ellipse? I'm struggling to relate these terms. Thanks for your help!

Answer 1

For convenience, I'll call the distance between the foci $\ell$.

The major axis goes through both foci; if you have the length constant $d$, the two lines it makes covers the distance between the foci once and the distance from one focus to the end of the ellipse twice. --Which actually means that it covers the semimajor axis (which is distance from center to end, passing through the focus) twice, so the major axis has length of exactly $d$.

The minor axis is slightly harder, being the perpendicular bisector of the major axis. The extreme point is distance $d/2$ from each focus, and is the hypotenuse of a right triangle that has one base that is the distance between one focus and the center $\ell/2$. ...conveniently, this means that the semiminor axis has length $\sqrt{(d/2)^2-(\ell/2)^2} = \sqrt{d^2-\ell^2}/2$, so the minor axis has length $\sqrt{d^2-\ell^2}$.

Answer 2

Ellipse property

$$ b^2 = a^2 - c^2 $$

$$ (2b)^2 = (2a)^2 - (2c)^2 $$

Note that since

$$ majoraxis=2a= d_1+d_2 ,$$

$$[minoraxis^2 =(2 a)^2 - (inter-focal distance)^2 ]$$

So

$$ minoraxis^2 = (d_1+d_2)^2 - (inter-focal distance)^2 $$

$$ minoraxis = \sqrt{(d_1+d_2)^2 - (inter-focal distance)^2}.$$