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It seems well-known that the system of conics given by $\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$ for $c>0$ fixed and $a \in (0,c)\cup(c,\infty)$ varying is orthogonal: whenever two of these curves intersect, they do so at a right angle. Does anyone know a good elementary proof of this? I.e. no complex analysis, no physics... something the ancients would've appreciated. I am looking to explain this to a sharp 13-year-old...

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The standard proof isn't complex analysis or physics, but it does use a bit of differential calculus. This is exercise 20 on page 510 of Apostol's Calculus book (at the end of Chapter 13, if editions vary) –  Charles Siegel Dec 8 '09 at 16:44

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As long as you can get an elementary proof of the fact that one of the families consists of ellipses with foci $A=(-c,0)$ and $B=(c,0)$ and the other consists of hyperbolas with the same foci, you can say that for any intersection point $P$ the angle between the lines $PA$ and $PB$ is dissected by the tangent to either curve - otherwise moving on the tangent would cause a first-order error in the sum $|PA|+|PB|$ or the difference $|PA|-|PB|$, respectively. Hence the two tangents are just the two angle bisectors of a pair of lines, and are thus orthogonal.

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Beautiful! Thanks a lot. –  David Hansen Dec 8 '09 at 16:56

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