Reflection Of Conic Section About A Line

Question

If a certain conic section $$ ax^2+2hxy+by^2+2gx+2fy+c=0 $$ is reflected about any line $y=mx+n$ what will be its new equation?

Answer 1

In general, the equation of the transformed conic is what you get when you apply the inverse transformation to a generic point $(x,y)$ and plug the result back into the original equation. That's because the image of the point (i.e. $(x,y)$) lies on the image of the conic (i.e. new equation) if and only if the preimage of the point (i.e. inverse transformation applied to image) lies on the preimage of the conic (i.e. original equation).

In the case of a reflection, things are fairly simple: the reflection is its own inverse, so simply apply it to a point and plug that into the equation. The line $y=mx+n$ or equivalently $mx-1y+n=0$ has normal vector $(m,-1)$. So the reflected point has coordinates

$$\begin{pmatrix}x\\y\end{pmatrix}-2\frac{mx-y+n}{m^2+1}\begin{pmatrix}m\\-1\end{pmatrix}$$

as various sources around the net will tell you. You can even verify that the midpoint lies on the line itself, as it should:

$$m\left(x-\frac{mx-y+n}{m^2+1}m\right) - \left(y+\frac{mx-y+n}{m^2+1}\right) + n = 0$$

You might want to scale your equation by $(m^2+1)^2$ to avoid all those fractions. You get the new coefficients

\begin{align*} a' &= m^{4} a - 4 m^{3} h - 2 m^{2} a + 4 m^{2} b + 4 m h + a \\ h' &= - m^{4} h - 2 m^{3} a + 2 m^{3} b + 6 m^{2} h + 2 m a - 2 m b - h \\ b' &= m^{4} b + 4 m^{3} h + 4 m^{2} a - 2 m^{2} b - 4 m h + b \\ g' &= 2 m^{3} n a - m^{4} g - 6 m^{2} n h + 2 m^{3} f - 2 m n a + 4 m n b + 2 n h + 2 m f + g \\ f' &= -2 m^{3} n h + m^{4} f - 4 m^{2} n a + 2 m^{2} n b + 2 m^{3} g + 6 m n h - 2 n b + 2 m g - f \\ c' &= 4 m^{2} n^{2} a - 4 m^{3} n g + m^{4} c - 8 m n^{2} h + 4 m^{2} n f + 4 n^{2} b - 4 m n g + 2 m^{2} c + 4 n f + c \end{align*}