Finding the point closest to the origin using euclidean norm

Question

I have the locus equation of some curve obtained using p norm as 2*x^2+4*x*y+3*y^2=1. How do I find the points closest to origin using Euclidean norm.

Answer 1

First solution:

Use polar coordinates $x=r\cos(\theta),y=r\sin(\theta)$ to convert the equation

$$\tag{0}f(x,y)-1=2x^2+4xy+3y^2$$

(of an ellipse, see graphics below) into:

$$\tag{1}r^2(2\cos(\theta)^2+4\sin(\theta)\cos(\theta)+3\sin(\theta)^2)=1$$

Making use of formulas

$$2\sin(\theta)\cos(\theta)=\sin(2\theta), \ \ \cos(\theta)^2=\frac{1}{2}(1+\cos(2a)), \ \ \sin(\theta)^2=\frac{1}{2}(1-\cos(2a)),$$

(1) can be transformed into:

$$\tag{2}r^2(1+\cos(2\theta)+2\sin(2\theta)+\frac{3}{2}(1-\cos(2\theta)))=1$$

$$\tag{3}\dfrac{r^2}{2}(5-\cos(2\theta)+4\sin(2\theta))=1$$

The minimization of $\dfrac{r^2}{2}$ is equivalent to the maximization of expression:

$$\tag{4}f(\theta):=5-\cos(2\theta)+4\sin(2\theta)$$

that can be written under the following form involving a dot product:

$$\tag{5}f(\theta):=5+\vec{U}.\vec{V} \ \ \text{with} \ \ \vec{U}=\binom{-1}{\ \ 4}, \vec{V}=\binom{\cos(2\theta)}{\sin(2\theta)}.$$

Its maximum occurs for a value of $\theta$ such that the dot product is maximized. It is well known that this maximization occurs when

$$\tag{6}\vec{U}=\lambda \vec{V} \ \ \text{with} \ \ \lambda>0$$

Taking norms on both sides of (6), we get $\|\vec{U}\|=\lambda \|\vec{V}\|$, i.e., $\lambda = \sqrt{17}.$

Thus the maximum value of $f(\theta)$ is $5+\sqrt{17}.$

Plugging this into (4) gives

$$\dfrac{5+\sqrt{17}}{2}r^2=1 \ \ \Leftrightarrow \ \ r=\frac{1}{2}\sqrt{5-\sqrt{17}}\approx 0.468213...$$

Having the shortest distance $r$, it is easy to see that the closest points to the origin are points of the form $(x,y)=(x,x)$ with $\sqrt{x^2+x^2}=r.$

Thus $x=\pm\dfrac{r}{\sqrt{2}}$.

2nd solution:

If you have seen Lagrange multipliers, it suffices to solve the system obtained by setting $g(x,y)=x^2+y^2$ and grouping the two equations $\vec{grad}f=\lambda \vec{grad}g$ with the equation of the conic section in this way:

$$\cases{4x + 4y = \lambda x \\ 4x + 6y = \lambda y \\ 2 x^2 + 4x y + 3 y^2 = 1}$$

which will give you at once the solutions. Of course, you will have to select those which provide a minimum from those that generate a maximum.

On the graphics below, one finds the conic section and the circle centered in O with radius $r$.