Question
The point on the parabola y=x^2 that is closest to the point (2,1/2) is (___,___). The distance between the two points is _______.
Answers
Steve
The distance z from (2,1/2) to (x,x^2) is
z = √[(x-2)^2 + (x^2 - 1/2)^2]
= 1/2 √(4x^4-16x+17)
dz/dx = 4(x^3-1)/√(4x^4-16x+17)
dz/dx=0 at x=1
z(1) = √5/2
Or, consider the line through (x,x^2) and (2,1/2). It has slope
(1/2 - x^2)/(2-x)
The tangent line to y=x^2 has slope 2x. So, we want the two lines to be perpendicular. That means
(1/2 - x^2)/(2-x) * 2x = -1
x=1 as above
z = √[(x-2)^2 + (x^2 - 1/2)^2]
= 1/2 √(4x^4-16x+17)
dz/dx = 4(x^3-1)/√(4x^4-16x+17)
dz/dx=0 at x=1
z(1) = √5/2
Or, consider the line through (x,x^2) and (2,1/2). It has slope
(1/2 - x^2)/(2-x)
The tangent line to y=x^2 has slope 2x. So, we want the two lines to be perpendicular. That means
(1/2 - x^2)/(2-x) * 2x = -1
x=1 as above