Question

Find the smallest distance of the point 0
c.From the parabola y=x^2

Answer 1

Step-by-step explanation:

Let f(t):=(t−0)2+(t2−c)2. The necessary condition for the minimum of f is

ddt{(t−0)2+(t2−c)2}=2t+4t(t2−c)=0,

which yields

t1=0,t2,3=±c−1/2−−−−−−√.

However, t2,3 are real only for c≥1/2. In addition, the second order condition shows that when c≥1/2, f(t1) is a local maximum. So, the minimum distance squared is c2 when c<1/2 and c−1/4, otherwise.