# Slope of axes of a General Conic Section

A General Conic Section is given by the equation $$ax^2 + by^2 + 2hxy +2gx +2fy + c =0$$. Let the $$\theta$$ be the slope of one of its axes.

Prove that :

$$\tan 2\theta = \dfrac{2h}{a-b}$$

I tried to use the definition of the conic section by taking cases for various conics such as parabola, pair of straight lines, ellipse, hyperbola etc. but it became too tedious to calculate. Are there some properties with which this question can be simplified ?

Any help will be appreciated.
Thanks.

• What do you mean by "axes" ? May 13, 2016 at 13:51
• @DeNiSkA Axes is the plural of axis. I'm referring to the directrix and the line perpendicular to it through the center. May 13, 2016 at 14:04
• @DeNiSka is right to ask this question, and don't understand why you are joking about singular or plural: if you speak of a directrix it is not an axis in the commonly accepted sense where "axis" means axis of symmetry (an axis with respect to which the curve is its own symmetrical curve). May 13, 2016 at 14:23
• @Jean i really didn't knew what axis or axes is referring to here! I it is directrix then it's OK May 13, 2016 at 14:34
• @DeNiSkall All right, sorry for having overinterpreted your thoughts :) Nevertheless, I maintain for Henry that a directrix is not an axis. You can take it as an axis of coordinates, yes, but it is never an axis of symmetry. May 13, 2016 at 14:49

if you shift the origin to eliminate the linear terms the equation becomes $$rMr^T = C$$ where $r$ is the 2-vector $(x,y)$ and $M$ is the matrix

$$M = \begin{pmatrix} a & h \\ h & b \end{pmatrix}$$ and $C$ is a constant.

If the axes are rotated through an angle $\theta$ the matrix is diagonalized when the off-diagonal terms in $$\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} a & h \\ h & b \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$ are zero. For the top right entry this requirement translates to: $$(a\cos \theta+h\sin \theta)(-\sin \theta)+(-h \sin \theta+b\cos \theta)\cos \theta = 0$$ i.e. $$h(\cos^2\theta -\sin^2 \theta)= (a-b)\sin\theta\cos\theta$$ from which the stated result follows

• This wiki page details matrix representation. Sep 18, 2021 at 1:47

If we rotate the axes through $\theta$ to new axes $x',y'$ we have $x=x'\cos\theta-y'\sin\theta,y=x'\sin\theta+y'\cos\theta$. Hence the coefficient of $x'y'$ becomes $-2a\cos\theta\sin\theta$ (from the $ax^2$ term) plus $2b\cos\theta\sin\theta$ (from the $by^2$ term) plus $2h(\cos^2\theta-\sin^2\theta)$ (from the $2hxy$ term).

So the $x'y'$ term has zero coefficient if $(a-b)\sin2\theta=2h\cos2\theta$ or $\tan2\theta=\frac{2h}{a-b}$.

But zero $xy$ term is precisely the condition for the axes of the conic to be lined up with the coordinate axes. Hence the slope of one of its axes (in the original $xy$ coordinate system) is $\theta$ where $\tan2\theta=\frac{2h}{a-b}$.

• Note that xy=0 can also be due to it being a circle, in which case the tan is also undefined as a=b. (a=b also happens in parabolas and hyperbolas with axis and tangent at vertex/axes of slope $\pm1$, but that makes sense anyway. ) Sep 19, 2021 at 7:17