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 shortest distance between the parabola $y = x^2$ and straight line $y = x - 1$ occurs when the tangent to the parabola is parallel to the line, i.e. when their slopes are equal. Now, slope of tangent of parabola $y = x^2$ is
$$
m = \frac{dy}{dx} = 2x
$$
and the slope of the straight line $= 1$. Hence the shortest distance occurs at $x = 1/2$. The point on the parabola corresponding to $x = 1/2$ is $(1/2, 1/4)$. Hence, perpendicular distance of this point from the given straight line is:
$$
d = \frac{|0.5 - 0.25 - 1|}{\sqrt{2}} = \frac{3}{4\sqrt{2}}
$$
which is the shortest distance between the curves.
EDIT: This, I guess, isn't the answer you're looking for. But I find this approach much easier and shorter than using Lagrange multipliers. Hence just shared this technique.
