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.