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

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Answer 1 · 2015-08-16T11:35:40

Approximately $(0.58975, 0.34781)$ $(0.58975, 0.34781)$

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

$\sqrt{{(x - 1)}^{2} + {(y - 0)}^{2}} = \sqrt{x^{4} + x^{2} - 2 x + 1}$ $sqrt((x-1)^2+(y-0)^2) = sqrt(x^4+x^2-2x+1)$

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

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

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

$y = x^2 ~= 0.34781$

Which point on the parabola y=x^2y=x2y=x^2 is nearest to (1,0)?