Find the points on the curve $$ax^2 + 2bxy + ay^2 = c$$
where $c > b > a > 0$ and whose distance from the origin is minimum.
My approach:
$(x_1)^2+(y_1)^2=D^2$
Putting the values $ax_1^2 + 2bx_1y_1 + ay_1^2 = c$
$aD^2 + 2bx_1y_1 = c$
I am not able to convert this implicit function.
On
The solution of @Andrei is very neato, here we go with the messy one. We have the following problem
$$\min_{x,y} \; (\sqrt{x^2+y^2}) \\ \text{s.t.} \; ax^2+2bxy+dy^2=c$$
Now this problem clearly is not convex, but we convexify it like this
$$\min_{x,y} \; (x^2+y^2) \\ \text{s.t.} \; ax^2+2bxy+dy^2 \le c$$
Now we solve it using the dual (Lagrangian) $$\mathscr{L}(x,y,\lambda) = x^2+y^2+\lambda(ax^2+2bxy+dy^2-c)\\ \nabla \mathscr{L}(x,y,\lambda) = \begin{bmatrix} (2+2a\lambda)x+2b\lambda y \\ 2b\lambda x+(2+2a\lambda) y \end{bmatrix} $$ The slackness condition is $\lambda(ax^2+2bxy+dy^2-c)=0$, Now here we have to be carefull, if we assume $\lambda=0$ then we are allowing to get away from boundary and since we relaxed the problem to be convex, the answer will be $x=y=\lambda=0$ which clearly is not our solution. Therefore we go with $\lambda \neq 0$ and $ax^2+2bxy+dy^2-c = 0$. In that case, by solving $\nabla \mathscr{L} = 0$ we get the trivial solution $x=y=0$ which we know is not our answer, Thus the gradient equations must be degenerate and the determinant of system is equal to zero, we have $$(2+2a\lambda)^2-(2b\lambda)^2=0$$
Which gives us two solutions, either $\lambda = \frac{-1}{a+b}$ or $\lambda =\frac{1}{b-a}$, putting those is one of the equations of $\nabla \mathscr{L}$ gives us either $x=y$ or $x=-y$. Now putting this in the constraint gives $ax^2+2bx^2+dx^2 = c$ or $ax^2-2bx^2+dx^2 = c$ which obviously gives u the answers $x^2=y^2 = \frac{c}{a+d\pm 2b}$
Therefore the minimum value for $\sqrt{x^2+y^2}$ is $\sqrt{\frac{2c}{a+d\pm 2b}}$. To get the answer of our problem we set $d=a$ and get $\sqrt{\frac{c}{a \pm b}}$ (the plus or minus sign is for the sign of $b$)
On
Let $(p,q)$ be the point on the curve with the minimum distance to the origin. Then, the normal vector of the curve $ax^2 + 2bxy + ay^2 = c$ at $(p,q)$ matches the vector $(p,q)$, i.e.
$$\frac{2ap+2bq}{2bp+2aq} = \frac pq$$
which leads to $p=\pm q$. Plug it into $ap^2 + 2bpq + aq^2 = c$ to get the coordinates of the valid point at
$$p=q=\pm\sqrt{\frac c{2(a+b)}}$$
On
As the involved functions are homogeneous, making $y = \mu x$ we have to minimize
$$ d^2 = x^2+y^2 = x^2(1+\mu^2) $$
subject to
$$ x^2(a+2b\mu + a\mu^2) = c\Rightarrow x^2 = \frac{c}{a+2b\mu+a\mu^2} $$
so now we can cosider the unrestricted problem
$$ (d^*)^2=\min_{\mu}\left(\frac{c(1+\mu^2)}{a+2b\mu+a\mu^2}\right) $$
and after deriving to determine the stationary points we get the condition
$$ \mu^2-1=0\Rightarrow (d^*)^2 = \frac{c}{a+b}\Rightarrow d^* = \sqrt{ \frac{c}{a+b}} $$
You can try to use polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$. Then your curve becomes $$ar^2+2br^2\sin\theta\cos\theta=c$$You can get $$r^2=\frac c{a+b\sin(2\theta)}$$ The minimum value occurs when $\sin2\theta=1$ or $x=y$ and therefore $$r=\sqrt\frac c{a+b}$$
You can get the same result if you use the Lagrange multiplier method