# find the nearest point from the point (4,1) using the equation 2y=x^2 We are asked to minimize the distance from the point (4,1) to the curve 2y=x^2:

Use a point on the curve (x,x^2/2). Then we can use the distance formula to find the distance between these points:

d=sqrt((4-x)^2+(1-x^2/2)^2)

=sqrt(16-8x+x^2+1-x^2+x^4/4)

=sqrt(x^4/4-8x+17)

Since the sqrt function is increasing on its domain, in order...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

We are asked to minimize the distance from the point (4,1) to the curve 2y=x^2:

Use a point on the curve (x,x^2/2). Then we can use the distance formula to find the distance between these points:

d=sqrt((4-x)^2+(1-x^2/2)^2)

=sqrt(16-8x+x^2+1-x^2+x^4/4)

=sqrt(x^4/4-8x+17)

Since the sqrt function is increasing on its domain, in order to minimize the distance we need only minimize the radicand:

To minimize x^4/4-8x+17 we take the first derivative and set it equal to zero -- solving this equation gives us the critical points from which we can determine the minimum:

d/dx(x^4/4-8x+17)=x^3-8

x^3-8=0

x^3=8

The only real solution is x=2, thus the minimum distance occurs when x=2. The point on the curve is (2,2).

Approved by eNotes Editorial Team