# Diagonals of an ellipse

I have been watching Norman Wildberger’s videos on all things mathematics for about 10 years. To say that I have learned a lot is the understatement of the decade.

His most recent video is a recorded talk from July this year titled “How Chromogeometry transcends Klein’s Erlangen Program for Planar Geometries”. It is fascinating throughout but my interest was piqued when at 25:13 he starts talking about ellipses.

In my research on 3D computer vision, I have used ellipses as an approximation for 2D forms. I am interested in the (central) projection of ellipses and their 3D counterparts ellipsoids. Applying Wildberger’s rational trigonometry, I have for example derived a formula for the semi-diameter $P$ of the orthographic projection of an ellipse with semi-diameters $A$ and $B$, and with spread $s$ indicating the rotation of the major axis with respect to the viewing direction: $P = s A + (1-s) B$

The formula is very satisfying in its simplicity, but also turns out to be extremely useful in understanding the central projection of 3D shapes.

In his talk Wildberger analyses the ellipse from the perspective of Chromogeometry, his three colored geometries. I want to focus here on the blue geometry, the classic Euclidean geometry. Wildberger states (see his slide below) that for a point on the ellipse, the quadrances of the altitudes with the two diagonals sum up to a constant quadrance $K$.

When I was watching this, I had a hunch about which lines these diagonals are. I made a quick calculation with 3 points on the ellipse with semi-diameters $A$ and $B$ and found an expression for $K$.

Wildberger was asked at the end of the talk which lines form the diagonals. He answered that he does not know of a geometric construction, since he derived them algebraically.

I propose here that the lines through the corners of the bounding box of the ellipse are the diagonals (the blue lines in the diagram below). I will prove that for any point on the ellipse Wildberger’s equation holds with $K = \frac{2 A B}{A+B}$.

We start with the classical expression for the ellipse at the origin with the major axis aligned with the x-axis: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Of course our quadrances $A$ and $B$ are the semi-axes $a$ and $b$ squared. The corners of the bounding box are the 4 points $(\pm a, \pm b)$. The diagonals go through the origin and the 2 pairs of opposite corner points respectively. The expression for the diagonal lines $l_1$ and $l_2$ are: $l_1: b x - a y = 0$ $l_2: b x + a y = 0$

Now we use a formula for the coordinates of the foot $\bold f_1$ of the altitude of point $\bold p = (p_x, p_y)$ on the line $l_1$ (Wildberger 2005, p. 42): $\bold f_1=(\frac{a^2 p_x+a b p_y}{a^2+b^2},\frac{a b p_x+b^2 p_y}{a^2+b^2} )$

The quadrance $Q_1$ between the point on the ellipse $\bold p$ and the foot $\bold f_1$ is: $Q_1 = (p_x - \frac{a^2 p_x+a b p_y}{a^2+b^2})^2 + (p_y- \frac{a b p_x+b^2 p_y}{a^2+b^2})^2$ $= (\frac{a^2 p_x + b^2 p_x - a^2 p_x-a b p_y}{a^2+b^2})^2 + (\frac{a^2 p_y + b^2 p_y - a b p_x -b^2 p_y}{a^2+b^2})^2$ $= \frac{ b^2(b p_x - a p_y)^2+a^2(a p_y- b p_x)^2}{(a^2+b^2)^2}$ $= \frac{(a^2 + b^2)(b p_x - a p_y)^2}{(a^2+b^2)^2}$ $Q_1 = \frac{(b p_x -a p_y)^2}{a^2+b^2}$

Similarly for the quadrance $Q_2$ between the point on the ellipse $\bold p$ and the foot $\bold f_2$ $\bold f_2=(\frac{a^2 p_x-a b p_y}{a^2+b^2},\frac{-a b p_x+b^2 p_y}{a^2+b^2} )$

We get: $Q_2= \frac{(b p_x + a p_y)^2}{a^2+b^2}$

Now we add the two quadrances $Q_1$ and $Q_2$ and use the fact that the point $\bold p$ is on the ellipse. So, $K = Q_1+ Q_2 = \frac{(b p_x -a p_y)^2+(b p_x + a p_y)^2}{a^2+b^2}$ $= \frac{b^2 p_x^2 + a^2 p_y^2+ b^2 p_x^2 + a^2 p_y^2}{a^2+b^2}$ $= \frac{2 a^2 p_y^2+ 2 b^2 p_x^2}{a^2+b^2}$

Since $\frac{p_x^2}{a^2}+\frac{p_y^2}{b^2}=1$ we can use $p_x^2=a^2(1-\frac{p_y^2}{b^2})$ $K = 2\frac{ a^2 p_y^2+ b^2 a^2(1-\frac{p_y^2}{b^2})}{a^2+b^2}$ $=2 \frac{ a^2 p_y^2+ b^2 a^2- a^2 p_y^2}{a^2+b^2}$ $=2 \frac{ a^2 b^2 }{a^2+b^2}$

When we use $A=a^2$ and $B=b^2$ we can express the formula for K as: $K =\frac{ 2 A B }{A + B}$

If this holds up, this result can be qualified as truly simple, both geometrically and algebraically.