Find $\min \left((x-2)^2+y^2\right)$ $s.t. \>x^2\leq ky^2+1$, $x\geq 0$ Consider the problem $$\min \left((x-2)^2+y^2\right)$$ $$s.t. \>x^2\leq ky^2+1$$
$$x\geq 0$$
where $k \in \mathbb{R}$ is a parameter of the problem.
Determine the status of the point $(1,0)$ for each value of $k$. For what values of $k$ is the point $(1,0)$ is a KKT point, local min,global min.
Solution so far
$L(x,\lambda)=(x-2)^2+y^2+\lambda_1(x^2-ky^2-1)-\lambda_2 x$

*

*$2(x-2)+2 \lambda_1 x-\lambda_2=0$

*$2y-2k \lambda_1 y=0$

*$\lambda_1\geq 0, x^2\leq ky^2+1,\lambda_1 (x^2-ky^2-1)=0$

*$\lambda_2\geq 0, x\geq 0, -\lambda_2 x= 0$
case $\lambda_1 >0, \lambda_2=0$
The third condition implies $x^2-ky^2-1=0$ But since $\lambda_2=0$ we have $$2y-2\lambda_1 ky=2y(1-k\lambda_1)=0$$
So $y=0, 1=k\lambda_1$ and  $(x^2-1)=0 \implies x=1,-1$
Observe $2(1-2)+2\lambda_1=0 \implies \lambda_1=1 \implies k=1$
$@(1,0),k=1,\lambda=(1,0)$

You are looking for the closest point to the point $(2,0)$ in an area limited by the hyperbole $x^2-ky^2=1$ on the right side and the y axis on the left side.
The solutions of the minimisation problem are clearly on the hyperbole.
So $x^2-ky^2=1$, $\lambda_2=0$
If $y \neq 0$ then $k=\frac{1}{\lambda_1}$ and $x(1+\frac{1}{k}) = 2$
$$x=\frac{2k}{k+1}$$
$$y=\pm \sqrt{\frac{(\frac{2k}{k+1})^2-1}{k}}$$
$$y=\pm \sqrt{\frac{3k^2-2k-1}{k(k+1)^2}}$$
So $(1,0)$ is solution when $k$ is so that $1\le (\frac{2k}{k+1}-2)^2+\frac{3k^2-2k-1}{k(k+1)^2}=\frac{3k^2+2k-1}{k(k+1)^2} = \frac{k-1/3}{k(k+1)}$
$$\iff k^2+k\le k-\frac{1}{3}$$
So $(1,0)$ is never a solution
