1. ## parabola

Find the point P on the parabola y = x² closest to the point (3, 0).

2. ## Re: parabola

$L=d^2 = (x-3)^2 + (x^2-0)^2 = (x-3)^2 + x^4$

$\dfrac{dL}{dx} = 2(x-3) + 4x^3$

$\dfrac{dL}{dx} = 0$

$2(x-3)+4x^3=0$

$2(3-x) = 4x^3$

$3-x=2x^3$

This has 1 real solution

$x=1$

$\dfrac{d^2L}{dL^2} = \left . 2+12x^2\right |_{x=1} = 14>0$

So this is a minimum.

The point is $(1,1)$

3. ## Re: parabola

hello thank you the point p is: p(1 ; 5) non (1; 1)

4. ## Re: parabola

Originally Posted by dhiab
hello thank you the point p is: p(1 ; 5) non (1; 1)
That is absolutely impossible. The point $(1,5)$ is not even on the parabola.

5. ## Re: parabola

Originally Posted by Plato
The point (1,5) is not even..
Agree: 1 and 5 are odd numbers