Question

Find the point on the parabola y^2=x that is closest to the point (1,7?

Answer 1

let the point of contact be P(a,b)
let A be (1,7)
slope of AP = (b-7)/(a-1)

the closest point would be where the tangent at P is perpendicular to AP
2y dy/dx = 1
dy/dx = 1/(2y)
so at P, the slope of the tangent is 1/(2b)

so (b-7)/(a-1) = -2b
b-7 = -2ab + 2b
-b = -2ab + 7

2ab - b = 7
b(2a - 1) = 7
b = 7/(2a-1) **

but since (a,b) lies on y^2 = x
b^2 = a
subbing that into **
b = 7(2b^2 - 1)
2b^3 - b = 7
2b^3 - b - 7 = 0
nasty:
Wolfram said, b = 1.6279
then a = 2.6501

http://www.wolframalpha.com/input/?i=solve+b%5E3+-+b+-+7+%3D+0

or

let the distance be D, and (x,y) as the closest point
D^2 = (x-1)^2 + (y-7)^2
but x=y^2
D^2 = (y^2 - 1)^2 + (y-7)^2
2D dD/dy = 2(y^2 - 1)(2y) + 2(y-7)
= 4y^3 - 4y + 2y - 14
= 4y^3 - 2y - 14
dD/dy = (2y^3 - y - 7)/D
= 0 for a min of D
2y^3 - y - 7 = 0

notice I got the "same" equation, except my y was b, and my x was a

the closest point is (2.6501, 1.6279)