# Ellipse inscribed angles

On wikipedia in German, we find relations about two angles inscribed on parable and on hyperbole. The 4 points of the parabola $y = ax^2 + bx + c$ has the following property: $$\frac{(y_4-y_1)}{(x_4-x_1)}-\frac{(y_4-y_2)}{(x_4-x_2)}=\frac{(y_3-y_1)}{(x_3-x_1)}-\frac{(y_3-y_2)}{(x_3-x_2)}$$ http://de.wikipedia.org/wiki/Parabel_%28Mathematik%29#Peripheriewinkelsatz_f.C3.BCr_Parabeln

The 4 points of hyperbole $y = \frac {a} {x-b} + c$ have the following property (slightly modified) : $$\frac{(y_4-y_1)}{(x_4-x_1)}/\frac{(y_4-y_2)}{(x_4-x_2)}=\frac{(y_3-y_1)}{(x_3-x_1)}/\frac{(y_3-y_2)}{(x_3-x_2)}$$ http://de.wikipedia.org/wiki/Hyperbel_%28Mathematik%29#Peripheriewinkelsatz_f.C3.BCr_Hyperbeln

But I can not find a similar formula (simple) for the ellipse. Does it exist? I guess we obtain the formula by the cross-ratio?

• The relations given depend strongly on the specific forms of the conic equations. The parabola relation doesn't even hold when the parabola is reflected over the line $y=x$ (that is, when $x_i \leftrightarrow y_i$). The hyperbola relation allows this reflection, and rotations by multiples of $90^\circ$, but not arbitrary rotations. I suspect that part of the magic here involves the fact that $y=ax^2+bx+c$ has $y$ as a function of $x$, and that $y=\frac{a}{x-b}+c$ has $x$ and $y$ as functions of each other. Ellipse equations aren't ever functions of either coordinate; so, no (simple) magic? – Blue Dec 6 '14 at 1:20
• Ok, no magic, but from where are these relations? Cross-ratio? – Amorok Dec 6 '14 at 1:51
• I would expect cross-ratio-based relations to be coordinate-independent in a way that these aren't. These relations may have arisen from simply observing that slopes of chords on $y=x^2$ and $y=1/x$ reduce nicely; writing $m_{ij}$ for the slope between $(x_i,y_i)$ and $(x_j,y_j)$, we have $$y=x^2 \;\to\; m_{ij} = x_i+x_j \qquad\qquad y = \frac{1}{x} \;\to\; m_{ij} = \frac{-1}{x_i x_j}$$ and then it's pretty easy to get $m_{13}$, $m_{14}$, $m_{23}$, $m_{24}$ to cancel. The fact that $y=x^2$ and $y=1/x$ happen to represent conics could just be a coincidence. – Blue Dec 6 '14 at 2:29

These relations may have arisen from simply observing that slopes of chords on $y=x^2$ and $y=1/x$ reduce nicely: Writing $m_{ij}$ for the slope between $(x_i,y_i)$ and $(x_j,y_j)$, we have $$y = x^2 \;\to\; m_{ij} = x_i + x_j\qquad\qquad y = \frac{1}{x}\;\to\;m_{ij}=-\frac{1}{x_ix_j}$$ and then it's pretty easy to get $m_{13}$, $m_{14}$, $m_{23}$, $m_{24}$ to cancel (by addition and subtraction in the first case, and by multiplication and division in the second case).
The fact that $y=x^2$ and $y=1/x$ happen to represent conics could just be a coincidence.