the distance d is found by
d^2 = (1-x)^2 + y^2
= 1-2x+x^2+4x
= (1+x)^2
d = 1+x
this is obviously a minimum when x=0, so (a)
Determine the point on the curve y^2=4x that is closest to the point (1,0)
a) (0,0)
b) (1,2)
c) (0.5,sqrt of 2)
1 answer