How can I find the minimum distance between the origin $(0,0)$ and the curve $y=1-x^2$ using Lagrange multipliers?

Question

I used the curve as the constraint and the origin as the point but I am not sure if that is correct.

Answer 1

Let $(x,y)$ be a point on the curve $y = 1 - x^2$, then the distance $d = \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}$. We minimize $d^2 = x^2+y^2 = f(x,y)$ instead of $d$ as it is quicker.

$C: x^2-1+y= 0$. So:

$f_x = \alpha C_x \to 2x = 2\alpha x$

$f_y = \alpha C_y \to 2y = \alpha$

Thus: $x(1-\alpha) = 0 \to x = 0$ or $\alpha = 1$.

So: $x = 0 \to y = 1$,and $\alpha = 1 \to y = \dfrac{1}{2} \to x^2 = \dfrac{1}{2}$. So $x = \pm \sqrt{\dfrac{1}{2}}$

Thus the possible points are: $A = (0,1), B= \left(\sqrt{\dfrac{1}{2}},\dfrac{1}{2}\right), C=\left(-\sqrt{\dfrac{1}{2}},\dfrac{1}{2}\right)$.

Check that there are $2$ points which minimize $d^2$ and hence $d$, namely: $B$, and $C$, and the corresponding minimized distance $d$ is $\dfrac{\sqrt{3}}{2}$.

Answer 2

No Lagrange multipliers needed. Let's call the point on the parabola whose distance from the origin is the minimum $(m,1-m^2)$.

The vector from the origin to this point is of course $\langle m,1-m^2 \rangle$. The tangent vector at this point is $\langle 1,-2m \rangle$. If $(m,1-m^2$) is indeed the point on the parabola whose distance is minimum from the origin, these two vectors must be perpendicular.

$$\langle m,1-m^2 \rangle\cdot\langle 1,-2m \rangle=2m^3-m=0$$

The only solutions are $m=0$ or $m=\pm\frac{\sqrt{2}}{2}$. The distance between $(0,1)$ and the origin is obviously $1$. The distance between $\left(\frac{\sqrt{2}}{2},\frac{1}{2}\right)$ and the origin is $\frac{\sqrt{3}}{2}$. We thus conclude that the minimum distance is $\frac{\sqrt{3}}{2}$.