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| My National Review Online "Diary" column for July 2005 included the following brainteaser.
Parabolic Chord
Problem —————— Consider a unit parabola, which is to say, one having focus at (1,0) and directrix x + 1 = 0. The equation of this parabola is of course y^2 = 4x.
At every point of this parabola there is a tangent, and a normal through the point at right-angles to the tangent.
A quick sketch will convince you of the following true fact: The normal at any point of the parabola, except the point (0,0), meets the curve again at a second point.
Another way to say that is: Every normal (except one) is a chord of the parabola.
Question: For which points of the parabola does this chord have minimum length? I was hoping someone would have a neat solution to this, but no-one had. I myself did it by breaking rocks: Find equation of normal at (v^2/4,v), figure out where it intersects the parabola again, compute length, differentiate w.r.t. v, find zero of derivative. Solution: The chord from (2,2*Sqrt[2]) and (8,-4*Sqrt[2]) for a chord length of Sqrt[108].
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