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.

(In reply to solution by Daniel)

You err

.......

and this is minimized when
4k^2+1-(1/4)k^-2-(1/16)*k^-4=0
4k^6+   k^4    -(1/4)k^2-(1/16)=0                             *16
64k^6+  k^4  -4k^2-1=0
..................

while multiplying by 16 you have left one member  (k^4) unchanged

therefore the rest is wrong , although your solution followed the right track.

Please correct and explain way of solving - I suggest substitutingb  m=k^2 to get a cubic equation.

Posted by Ady TZIDON on 2009-12-02 14:43:34