Trouble with finding shortest distance between $y=x^2$ and $y=x-1$ with Lagrangian multipliers I've found another thread with a similar question, but none of the answers help with the specific part I'm stuck on. 
Just to make things simpler, I've used the square of the distance $f(x_1,x_2,y_1,y_2)=(x_1-x_2)^2+(y_1-y_2)^2$, and I have constraints $G_1=x_1^2-y_1$, and $G_2=x_2-y_2-1$.
Taking the gradients of $f$, $G_1$, and $G_2$, I have the system of equations
$2(x_1-x_2)=2\lambda_1x_1$
$2(y_1-y_2)=-\lambda_1$
$-2(x_1-x_2)=\lambda_2$
$-2(y_1-y_2)=-\lambda_2$
What I've done so far is, I've first observed that the immediate implication of the 2nd and 4th equations is that $\lambda_1=-\lambda_2$. Using this, I cancelled out the 1st and 3rd equations with the substituted value for $\lambda_2$, and got that $2\lambda_1x_1=\lambda_1\implies x_1=\frac{1}{2}$, and by constraint 1 that $y_1=\frac{1}{4}$. From here though, I'm not sure how to pin down the value of $\lambda$, and therefore determine $x_2$ and $y_2$.

the distance between the curves is given by $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ with $$y_1=x_1^2,y_2=x_2-1$$ with these equations we obtain
$$\sqrt{(x_2-x_1)^2+(x_2-1-x_1^2)^2}$$ now you must differentiate this with respect to $$x_1,x_2$$
