the algebraic interior-exterior theorem
Suppose that C is the projective conic f(x) = xTMx = 0.
Then P = [p] and Q = [q] both lie on the same side of C
if and only if
f(p) and f(q) have the same sign.
proof
Suppose first that P lies inside C.
Then L, the polar of P, lies entirely outside C.
Also, the polar has equation pTMx = 0, by Joachimsthal's formulae.
Let R = [r] lie on L, so pTMr = 0.
Note that, as M is symmetric, transposing this gives rTMp = 0.
Since P is inside C, the p-line PR cuts C twice.
Let A = [a] be one of the intersections. as A is on C, f(a) = 0.
As A is on PR, there exist constants λ, μ such that a = λp+μr.
Since A ≠ P, R, λ, μ ≠ 0.
We have 0 = f(a) = (λp+μr)TM(λp+μr).
Multiplying out the right hand side, we get λ2f(p)+μ2f(r),
since pTMr = rTMp = 0.
Thus λ2f(p)+μ2f(r) = 0.
Since λ2 and μ2 are positive,
f(p) and f(r) must have opposite sign.
If Q also lies inside C, we can let R be the intersection of the polars
of P and Q. As above f(p) and f(r) have opposite sign, as do f(q) and f(r).
Thus f(p) and f(q) have the same sign.
Now suppose that P and Q lie outside C.
Then the polar of P will cut C twice, so we can choose R on the polar
and inside C. As above f(p) and f(r) will have opposite sign.
Likewise, we choose S = [s] inside C so f(q) and f(s) have opposite sign.
Now R and S are inside C, so f(r) and f(s) have the same sign.
It follows that f(p) and f(q) have the same sign.
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