Further to this question Intersections of Tangent Lines to Parabola Different Proof (showing that there are at most two tangents from a point in a plane to a parabola), it struck me that I don't know an efficient proof that there are at most two tangents from a fixed point to a smooth convex curve in the plane. For this purpose a convex curve can usefully be thought of as the boundary of a convex set of points, and smooth as having a tangent at every point on the curve.
This ought to be possible using basic properties of convexity, rather than explicit computation. I can think of approaches that should work. But is there an efficient or canonical approach?
I believe I found a counterexample. Tell me what do you think about this
The curve is $y=x^4-6 x^3-35 x^2+132 x+160$, the point is $A(-7,10)$ and the four tangent lines are $$y=-3529.61 x-24697.3,y=-99.9887 x-689.921,\\y=-25.0034 x-165.024,y=30.4533 x+223.173$$
Edit
You are right. I missed the convexity issue. A convex curve, closed or not, lies in the same half plane with origin one of the tangent. Therefore if two tangents intersect in one point, the curve is included in the convex angle having sides the two tangents and any other line passing through that point intersects the curve is a simple intersection. (BTW: A convex curve cannot have self intersections)
Don't know if it's enough as proof.
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