Finding the shortest distance between an arbitrary point and a parabola I'm attempting to find the shortest distance between a point and a parabola. The point in question is (0,b), for any b, and the parabola that we are given is$\ y = x^2 $.
How would you approach the problem and find the shortest distance for any given b?
What about if the point was (0,0,b), and the equation was $\ z = x^2 + y^2$?

2D
We consider the distance between a query point $Q=(0,b)$ and some point $P(x) = (x, x^2)$ on the graph of the function. 
This leads to
$$
d(x) = d(Q, P(x)) = \sqrt{x^2 + (x^2 - b)^2}
$$
The distance to the graph is the minimum of those distances:
\begin{align}
d &= \min_{x \in \mathbb{R}} d(x) \\
  &= \min_{x \in \mathbb{R}} \lVert (x, x^2)) - (0,b)\rVert
\end{align}
So we look for local extrema via
$$
0 
= d'(x) 
= \frac{2x + 2(x^2 -b)\,2x}{2\sqrt{x^2 + (x^2 - b)^2}}
= \frac{2x^3+(1-2b)x}{\sqrt{x^2 + (x^2 - b)^2}}
$$
which happens for $x = 0$ or
$$
2x^2 = 2b - 1 \iff \\
x = \pm\sqrt{b - 1/2}
$$
which has real solutions for $b \ge 1/2$. The resulting distance for both of these solutions is 
$$
d 
= \sqrt{b - 1/2 + (b - 1/2 - b)^2} 
= \sqrt{b-1/4} 
= \sqrt{4b - 1 }/2
$$
while the distance for $x=0$ is between $(0,0)$ and $(0,b)$ thus $\lvert b \rvert$.
So the minimum distance is
$$
d = 
\begin{cases}
\min(\lvert b \rvert, \sqrt{4b - 1}/2 & \text{for } b > 1/2 \\
\lvert b \rvert  & \text{for } b \le 1/2
\end{cases} 
$$

(Large version)
3D
The above generalizes to three dimensions. Query point is $(0,0,b)$ and the points on the graph are $P(x,y) = (x, y, x^2 + y^2)$.
This leads to
$$
d(x,y) = d(Q, P(x,y)) = \sqrt{x^2 + y^2 + (x^2 + y^2 - b)^2}
$$
For local minima we look where the gradient vanishes:
\begin{align}
0 = \text{grad } d(x, y) 
&= \frac{1}{2\sqrt{x^2 + y^2 + (x^2 + y^2 - b)^2}}
(2x + 2(x^2 + y^2 - b)2x, 2y + 2(x^2 + y^2 - b)2y) \\
&= \frac{1}{\sqrt{x^2 + y^2 + (x^2 + y^2 - b)^2}}
(2x(x^2 + y^2) + (1-2b)x, 2y(x^2+y^2) + (1-2b)y)
\end{align}
This is the case for $(x,y) = (0,0)$ otherwise if
$$
2(x^2 + y^2) = 2b - 1 \iff \\
x^2 + y^2 = r^2 \wedge r = \sqrt{b - 1/2}
$$
which has real solutions for $b \ge 1/2$. The resulting distance for solutions on this circle in the $x$-$y$-plane with radius $r$ is:
$$
d = \sqrt{b - 1/2 + (b - 1/2 - b)^2} = \sqrt{4b-1}/2
$$
while the distance for $(x,y)=(0,0)$ is between $(0,0,0)$ and $(0,0,b)$ thus $\lvert b \rvert$.
So the minimum distance is again
$$
d = 
\begin{cases}
\min(\lvert b \rvert, \sqrt{4b - 1}/2 & \text{for } b > 1/2 \\
\lvert b \rvert  & \text{for } b \le 1/2
\end{cases} 
$$
