P(x,y) is an arbitrary point on the parabola y=x^2

Question

A= Express the distance "d" from "P" to the Point A(0,1) as a function of the y-coordinate of P

 

B= What is the minimum distance "d" 

Answer 1

d = √[(0-x)² + (1-y)²]    

 

Since y = x², x = ±√y

 

So, d = √[y + 1 - 2y + y²]

 

     d = √(y² - y + 1)

 

d is minimized when y² - y + 1 is minimized

 

     Minimum occurs at the vertex.  So, y = -(-1) / (2(1)) = 1/2

 

Minimum distance d = √[(1/2)² - (1/2) + 1] = √(3/4) = √3/2