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Given a general ellipse $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ where $B^2<4AC$, what are the major and minor axes of symmetry in the form $ax+by+c=0$?

It is possible of course to first work out the angle of rotation such that $xy,x,y$ terms disappear, in order to get an upright ellipse of the form $x^2/p^2+y^2/q^2=1$ and proceed from there. This may involve some messy trigonometric manipulations.

Could there be another approach perhaps, considering only quadratic/diophantine and linear equations?

Addendum

Here's a graphical implementation based on the answer by Ng Chung Tak.

Addendum 2

Based on the answers by amd and by Ng Chung Tak, the equations for the axes are

$$\color{red}{\left(y-\frac {2AE-BD}{B^2-4AC}\right)=\frac {C-A\pm \sqrt{(A-C)^2+B^2}}B\left(x-\frac {2CD-BE}{B^2-4AC}\right)}$$

Note that $$\frac{C-A\pm \sqrt{(A-C)^2+B^2}}B\cdot \color{lightgrey}{\frac {C-A\mp\sqrt{(A-C)^2+B^2}}{C-A\mp\sqrt{(A-C)^2+B^2}}}=-\frac B{C-A\mp\sqrt{(A-C)^2+B^2}}$$ i.e. it is equal to the negative of its own reciprocal. Hence the equations for the axes can also be written as

$$\color{red}{\left(x-\frac {2CD-BE}{B^2-4AC}\right)=-\frac {C-A\mp \sqrt{(A-C)^2+B^2}}B\left(y-\frac {2AE-BD}{B^2-4AC}\right)}$$

hence the two symmetrical anti-symmetrical forms for the axes. Here's the graphical implementation.

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    $\begingroup$ Hint: there’s a way to find a rotation that eliminates the cross term without using any trigonometric functions. Do you know about eigenvalues and eigenvectors of matrices? $\endgroup$ – amd Dec 4 '16 at 5:27
  • $\begingroup$ @amd - Yes of course. $\bf {Ax}=\lambda \bf x$. That would be an interesting approach. Will think about it. But the intention of the question is to see if it can be deduced without considering rotation. $\endgroup$ – hypergeometric Dec 4 '16 at 6:24
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    $\begingroup$ The eigenvectors of the matrix of a quadratic form, which is a symmetric real matrix, define its principal axes. That they happen to be orthogonal and thus allow you to define an orthogonal change-of-basis matrix—a rotation—is an extra benefit, but those principal axes can be found without reference to any rotation. That’ll give you the directions of the principal axes of the conic and, presumably, you have some other way of finding the center (for an ellipse or hyperbola, it’s the point at which both partial derivatives vanish). $\endgroup$ – amd Dec 4 '16 at 19:48
  • $\begingroup$ @ccorn - Thanks. Edited accordingly. $\endgroup$ – hypergeometric Dec 6 '16 at 11:31
  • $\begingroup$ @ccorn - Yup, that's right. Shd be OK now. $\endgroup$ – hypergeometric Dec 6 '16 at 11:52
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Let $f(x,y)=Ax^2+Bxy+Cy^2+Dx+Ey+F$. This function represents a quadratic surface, specifically an elliptic paraboloid in our case of $B^2<4AC$. The level curves of this surface form a family of concentric ellipses. The center of this family coincides with an extremum of the surface, so we can find it by solving the equation $\nabla F=0$. This yields $$\mathbf p=\left({BE-2CD\over4AC-B^2},{BD-2AE\over4AC-B^2}\right)$$ for the center of the family of ellipses. Alternatively, one can look for a translation that makes the linear terms vanish, which leads to the same equations, but with a bit more work.

Translating the paraboloid clearly doesn’t change its orientation, so we need examine only the quadratic part $Q(x,y)=Ax^2+Bxy+Cy^2$, which is unchanged by translations, to determine the directions of the principal axes. A fairly simple way to do this is to compute the eigenvectors of the matrix $\bigl(\begin{smallmatrix}A&B/2\\B/2&C\end{smallmatrix}\bigr)$ of this quadratic form. It’s a symmetric real matrix, so its eigenspaces are orthogonal and correspond to the ellipses’ axes: the minor axis to the eigenspace of the greater of the two eigenvalues, the major axis to that of the lesser. A bit of work produces the eigenvalues $$\frac12\left(A+C\pm\sqrt{(A-C)^2+B^2}\right)$$ with corresponding eigenvectors $$\left(A-C\pm\sqrt{(A-C)^2+B^2},B\right).$$You can verify that these vectors are indeed orthogonal.

Another way to compute these direction vectors is to find the extrema of $Q$ restricted to the unit circle. As with the eigenvalues, the major axis will lie in the direction of the local minima, the minor axis in the direction of the maxima. This can be solved via Lagrange multipliers (you will find that they are equal to the eigenvalues of $Q$) or by pulling $Q$ back by $\phi:t\mapsto(\cos t,\sin t)$, though this last method might lead to “messy” trigonometric manipulations.

With the center of the ellipse and direction vectors of the axes in hand, it’s a simple matter to construct the equations of the axis lines.

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  • $\begingroup$ Thanks for your answer. That's very clear indeed. (+1). As compared to the related question for a parabola, the final form doesn't appear to be as neat, but perhaps that's the nature of things. Would be interesting to see if it's possible to derive this answer using a similar method use in the solution here. $\endgroup$ – hypergeometric Dec 5 '16 at 13:32
  • $\begingroup$ The final form doesn't seem quite symmetrical on a stand-alone basis, although by using surds it can be transformed to a symmetrically asymmetrical (!) form (i.e. long/short coeff for $x,y$ instead of short/long). Is there a nice symmetrical form which can be expressed in one equation? $\endgroup$ – hypergeometric Dec 5 '16 at 13:55
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The principal axes are:

$$\left( y-\frac{2AE-BD}{B^2-4AC} \right)= \frac{C-A \color{red}{\pm} \sqrt{(A-C)^2+B^2}}{B} \left( x-\frac{2CD-BE}{B^2-4AC} \right)$$

Also refer to another answer.

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  • $\begingroup$ Thanks for your answer (+1). Can it be derived without considering the angle of rotation, but purely using algebraic means, e.g. similar to the answer here? $\endgroup$ – hypergeometric Dec 5 '16 at 8:11
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Idea for solution using multivariable calculus:

(1) Find the max an min of the functions $H(x,y) = x$ , $V(x,y) = y$ with the restriction $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ (easy with Lagrange multipliers). This gives a "bounding box" for the ellipse. The center of the box will be the center of the ellipse.

(2) Find the max and min distances of $(x,y)$ to the center of the ellipse with the restriction $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

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  • $\begingroup$ Thanks for your answer. (+1) $\endgroup$ – hypergeometric Dec 5 '16 at 13:29
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The trigonometric manipulations are not messy.

After centering, the equation is, in polar coordinates,

$$\rho^2=-\frac{F'}{A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta}.$$

The denominator $$A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta=\frac12\left((A-C)\cos\frac\theta2+B\sin\frac\theta2\right)$$ achieves its extrema when

$$(A-C)\sin\frac\theta2=B\cos\frac\theta2,$$ which is elementary.

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  • $\begingroup$ Thanks for your answer. (+1) $\endgroup$ – hypergeometric Dec 5 '16 at 13:29

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