Suppose I have an ellipse centered at the origin, and two points on its perimeter which are not antipodal to one another (i.e. not negative to each other as vectors in $\mathbb R^2$). How can I draw the whole ellipse?
How to draw an ellipse using its center and two points on its perimeter
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0@HenningMakholm Like in the example animation here http://en.wikipedia.org/wiki/Singular_value_decomposition – 2012-06-05
2 Answers
The Wikipedia animation you point to show an ellipse being generated by two vectors $\mathbf a$ and $\mathbf b$ via the parametric equation $ t \mapsto \cos(t)\cdot \mathbf a + \sin(t)\cdot \mathbf b $ with $t$ ranging from $0$ to $2\pi$, say.
Getting an equation for the ellipse is not as simple to describe. What you need there is to express an arbitrary point in the (non-rectangular) coordinate system that has $\mathbf a$ and $\mathbf b$ as basis vectors. Then the equation for (say) the closed ellipse is $ \{ v \mid p^2+q^2\le 1 \text{ where } v=p\mathbf a+q\mathbf b \}$
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1@Mahoni: For plotting, use the parametric function in my first equation. There you don't need to find the $p$ and $q$. – 2012-06-05
Hint: Write vectors in polar coordinates $(r,\theta )$ and use the polar coordinate ellipse equation for both vector and find (a and e) for the ellipse with the origin at one focus, with the angular coordinate $\theta = 0 $ still measured from the major axis,
the ellipse's equation is $r(\theta)=\frac{a (1-e^{2})}{1 \pm e \cos\theta} $ .
The eccentricity of the ellipse (commonly denoted as either e or $\epsilon$) is $e=\varepsilon=\sqrt{\frac{a^2-b^2}{a^2}} =\sqrt{1-\left(\frac{b}{a}\right)^2} =f/a$
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