Which point on the parabola #y=x^2# is nearest to (1,0)?

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Answer 1 · 2015-08-16T11:35:40

Approximately #(0.58975, 0.34781)#

I'm just going to solve this by the first method that comes to me, rather than trying to use any special geometric properties of parabolas.

If #(x, y)# is a point on the parabola, then the distance between #(x, y)# and #(1, 0)# is:

#sqrt((x-1)^2+(y-0)^2) = sqrt(x^4+x^2-2x+1)#

To minimize this, we want to minimize #f(x) = x^4+x^2-2x+1#

The minimum will occur at a zero of:

#f'(x) = 4x^3+2x-2 = 2(2x^3+x-1)#

graph{2x^3+x-1 [-10, 10, -5, 5]}

Using Cardano's method, find

#x = root(3)(1/4 + sqrt(87)/36) + root(3)(1/4 - sqrt(87)/36) ~= 0.58975#

#y = x^2 ~= 0.34781#