Set up the system of equations required according to Lagrange in order to minimize the square of the distance between P1 and P2.
$${M_{1}=\lbrace(x,y)\in R^2} \mid x^2+\frac{9}{4}(y-1)^2=9\rbrace$$
$${M_{2}=\lbrace{(x,y)\in R^2} \mid x^2+y^2 = \frac{1}{4}}\rbrace $$
Does the system of equations from above admit a solution at $(P_{1},P_{2})=(x_{1},y_{1},x_{2},y_{2})=(3,1,\frac{1}{\sqrt2},\frac{1}{\sqrt2})$?
That system of equations does not have that solution for the minimal square of distance, since your suggested point $(\frac 1{\sqrt 2},\frac 1{\sqrt 2})$ does not lie on curve $M_2$. For that point, $x^2+y^2$ is $1$, not $\frac 14$.
We do not even need Lagrange multipliers for this. Minimizing the distance between $P_1$ and $P_2$ will do the same as your problem, and the diagram clearly shows which points will do that.
So the correct solution is $(0,-1,0,-\frac 12)$.