Bonsoir,
Soit P=(0,1)
M=(x,y) un point de la courbe y=x².
[tex]PM=\sqrt{(x-0)^2+(x^2-1)^2)}=\sqrt{x^2+(x^2-1)^2)}\\0=\dfrac{dPM}{dx}=\dfrac{2x+2(x^2-1)*2x}{2\sqrt{x^2+(x^2-1)^2} }\\\\=\dfrac{2x(1+2(x^2-1))}{2\sqrt{x^2+(x^2-1)^2} }\\\\2(x^2-1)=-1\\x^2=\dfrac{1}{2}\\\\\boxed{x=\pm\dfrac{\sqrt{2}}{2}}\\\\Sol=\{(-\dfrac{\sqrt{2}}{2},\dfrac{1}{2}),(\dfrac{\sqrt{2}}{2},\dfrac{1}{2})\}[/tex]
