# Create Ellipse From Eccentricity And Semi-Minor Axis

So I am given the eccentricity of an ellipse and the radius semi-minor axis as well as the center of the ellipse. So in the example below we know the center of the ellipse is at ( 0, 0 ) and the radius of the semi-minor axis is 10. Let's say for the sake of the example the eccentricity is 0.75.

So my question is is it possible to find the points of the foci and the radius of the semi-major axis? Thaks!

• Since rotation preserves the eccentricity and the length of the semiminor axis, one also needs the orientation of the ellipse (e.g., the direction of the semiminor axis, as indicated in the picture) to specify the ellipse. – Travis Jul 31 '15 at 18:46
• @Travis, Let's say the orientation is always like the example. – CMilby Jul 31 '15 at 21:45

## 2 Answers

Hint The distance $f$ from the center to either focus is related to the length $a$ of the semimajor axis and the eccentricity $e$ by $$f = e a,$$ and the $f$, $a$, and the length $b$ of the semiminor axis are related by $$a^2 - b^2 = f^2.$$

• That makes sense! Thank you! – CMilby Aug 1 '15 at 2:01
• You're welcome, I'm glad you found it useful. – Travis Aug 1 '15 at 2:07

$b^2=a*2(1-e^2)$ gives the relation between eccentricity $e$ , major semiaxis ${a}$ and minor semiaxis ${b}$

In the example above you will have $$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$$ since the majot semiaxis is coincident with the $y$ axis