Posted by Elizabeth on Wednesday, July 18, 2012 at 5:51pm.
From the given point P0(x0,y0)=(2,1/2), we draw a line to join a point on the curve
f(x)=x^2 at P(x,y).
The distance P-P0 will be shortest when P-P0 is perpendicular to the tangent of f(x) at P.
The slope of P-P0 is given by:
m1=(y0-y)/(x0-x)
=(y0-x^2)/(x0-x) since P is on curve f(x)=x^2.
The tangent of the curve f(x) at P is given by
m2=f'(x)=2x
Perpendicularity requires that m1*m2=-1, or
[(y0-x^2)/(x0-x)]*(2x)=-1
which simplifies to a cubic equation
x³-1=0 which factorizes easily to
(x-1)(x²+x+1)=0
The first factor gives x=1, and the second factor results in two complex roots which do not concern us.
Therefore the required point is P(1,1), or P(1,1) is the point on f(x) that is closest to the given point (1, 1/2).
Thanks!
You're welcome!