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A Normal and a Parabola (Posted on 2009-12-02) Difficulty: 3 of 5
Choose any point (k,k^2) with k>0 on the parabola y=x^2. Draw the normal line to the parabola at that point. Then there is a closed region defined by the parabola and the line. Find the value of k so the area of the region is minimized.

Note: A normal line is a line perpendicular to a tangent at the point of tangency.

See The Solution Submitted by Brian Smith    
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solution | Comment 1 of 4

at point (k,k^2) the slope of the tangent is 2k so the slope of the normal is -1/(2k).  thus the equation of the normal line is
to find the other point of intersection equate it with y=x^2
x=k which we already knew
so we want to find the integral of
(-1/(2k))(x-k)+k^2-x^2 dx from x=-(1+2k^2)/(2k) to k
and this is easily found to be
and this is minimized when
4k^2+1-(1/4)*k^-2-(1/16)*k^-4=0    *16k^4
k=1/2 k=-1/2 k=i/2 k=-i/2
since we want positive real k then we are left with k=1/2
and this gives minimum area of 4/3

Edited on December 2, 2009, 10:02 pm
  Posted by Daniel on 2009-12-02 12:13:35

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