In Extraordinary Conics: The Most Difficult Math Problem I Ever Solved, 7:36, there are some formulas shown that apply for conjugate axis vectors in general:
- Area: $\pi|A\times B|$
- $C^2$ Invariant: $|A|^2 + |B|^2$
- Inside test: $|(P-C)\times A|^2 + |(P-C)\times B|^2 < |A \times B|^2$
- Tangent test: $|R\times A|^2 + |R\times B|^2 = |R\times (P - C)|^2$
where, for a real ellipse, $A$ and $B$ are one pair of conjugate axis vectors, and $C$ is the position vector of its center. $P$ is an arbitrary point in the plane, and $R$ is a vector. The "line" in the Tangent Test is defined as the line $C + kP$, where $k \in \mathbb R$.
My first instinct is to use the fact that an ellipse is just an affine transform of a circle, and hope that some kind of tensor transform would show those are scalars, and thus transform invariants. However that obviously doesn't work, especially in the case of $|A|^2 + |B|^2$, where $|A|^2$ is decidedly not invariant under affine transforms.
So, under what viewpoint would these formulas be obvious?
I don't want some ad hoc viewpoint, but a symmetric viewpoint, something in the spirit of Klein's program. Basically, I want a viewpoint whereby "an ellipse equipped with a pair of conjugate axis vectors" is just a symmetric image of "a circle equipped with a pair of perpendicular radius vectors", and the formulas are invariants of the symmetry.
The affine transform does not leave the formulas invariant, so it can't be used as it is, although I suspect that it is in fact the correct viewpoint. It's just that the formulas are not written in the right form. If in the right form, they would be invariant.
So far I found that the easiest way to show them is:
- For area being $\pi|A\times B|$, shear parallel to $B$, until $A$ is perpendicular to $B$.
- For the $C^2$ invariant $|A|^2 + |B|^2$, take the principal axis vectors $a, b$, then write $A = a\cos\theta + b \sin\theta$, $B = -a\sin\theta + b \cos\theta$, then calculate, using $a \cdot b = 0$.
- For the Inside Test, write $P = r(A\cos\theta + C, B\sin\theta + C)$, where $r > 0$, then plug in and calculate, to find that the inequality holds iff $ r < 1$.
- For the Tangent Test, write $P = (A\cos\theta + C, B\sin\theta + C) + k R$ to check that it works. Then note that if we displace $P$, the left side stays the same, but the right side changes linearly with displacement of $P$, with kernel spanned by $R$.