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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1In which way do your vectors describe the ellipse? – 2012-06-05
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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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0Okay so the equation is a set over the coordinates, isn't it? But how do I obtain the p and q values? – 2012-06-05
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0That depends on how your $v$'s are given, but basically it's a question of solving two linear equations in two unknowns for each $v$. One can do this symbolically for an arbitrary $v=(v_1,v_2)$, which produces $p$ and $q$ as certain linear combinations as $v_1$ and $v_2$. The coefficients in the solution are the entries of the inverse of the matrix whose rows are the coordinates of $\mathbf a$ and $\mathbf b$ in _standard_ coordinates. – 2012-06-05
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0This seems like the result are an arbitrary number of points, I thought there is something like a function which could be passed to my plotting program, plotting the ellipse. – 2012-06-05
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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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