## “Parabolas of Apollonius?” Spoiler

This is an answer to the question posed in Parabolas of Apollonius? which was inspired by Maria Monk’s blogpost on the Circles of Apollonius. Under the map $z \mapsto z^2$, vertical lines in the complex plane map to the blue parabolas (given by $x = a - \frac{1}{4a}y^2$ for $a > 0$) and horizontal lines map to the red parabolas (given by $x = a - \frac{1}{4a}y^2$ for $a < 0$). Details are in the next paragraph. Because the complex function $f(z) = z^2$ is differentiable, it is conformal. Therefore, if you take one red and one blue parabola, one of their intersections is perpendicular by conformality and the other because of symmetry. (It’s also possible to use conformality to explain perpendicularity at every intersection because for every parabola, there are two lines that are mapped onto it. So if you look at the image of a horizontal and vertical line, there intersection will correspond to one of the intersections between the blue and red parabola while the other intersection will correspond to the intersection between the same horizontal and the other vertical that maps to the same red parabola (or the same vertical and the other horizontal that maps to the same blue parabola).)

For details, note that $(m+ni)^2 = m^2 - n^2 + 2mni$, where $i = \sqrt{-1}$. If we let $x$ be the real part and $y$ be the imaginary part of this expression, we find that $x = m^2 - n ^2 = m^2 - (\frac{y}{2m})^2 = (\frac{y}{2n})^2 - n^2.$ 