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 $\ z = x^2 + y^2$?
We seek $x\in\Bbb{R}$ to minimize the distance between $(0,b)$ and $(x,x^{2})$. This amounts to minimizing the value of $$\sqrt{(x-0)^{2}+(x^{2}-b)^{2}},$$ and this gives the same value as the problem of minimizing $$x^{2}+(x^{2}-b)^{2}.$$ To do this, we set the derivative equal to zero and solve: \begin{eqnarray*} 2x + 4x(x^{2}-b) & = & 0,\\ \implies 2x[2x^{2}-2b+1]=0,\\ \implies x=0\:\text{ or }\:x^{2}=b-\tfrac{1}{2}. \end{eqnarray*} Of course, the latter is only valid for $b\geq 1/2$.
Therefore if $b<1/2$, then the minimum distance is $\sqrt{0+(0-b)^{2}}=|b|$. If $b\geq 1/2$, then the minimum distance is either $|b|$ or is $\sqrt{b-\tfrac{1}{4}}$, whichever is smaller, which by inspection is the latter.
As a check, note that $\lim_{b\to1/2^{-}}|b| = 1/2 = \lim_{b\to1/2^{+}}\sqrt{b-\frac{1}{4}}$, which satisfies our intuitions.
It is worth asking what the relevance is of the solution $x=0$ in the case $b\geq1/2$. Some thought (and perhaps a good diagram) shows that $x=0$ in fact gives a local maximum for the distance.
In $\mathbb{R}^{3}$, when we have $z=x^{2}+y^{2}$ and $(0,0,b)$, we can reduce the problem to the two dimensional one due to the rotational symmetry of the paraboloid and the fact that the point $(0,0,b)$ lies on the axis of symmetry. The major difference between the two situations is that instead of two points giving the same minimum distance ($x=\pm\sqrt{b-\frac{1}{2}}$ above), we instead have infinitely many (in fact, the points of minimum distance will form a circle, the intersection between some cone with vertex at $(0,0,b)$ and the paraboloid $z=x^{2}+y^{2}$).
Let the point on the parabola be $(x,x^2)$, so that the distance to the point is
$$d^2(x)=x^2+(b-x^2)^2.$$
To find the minimum value of this expression, derive on $x$ and equate to zero:
$$2x-4x(b-x^2)=0.$$
The possible solutions are $$x=0\lor x=\pm\sqrt{\frac12-b}.$$
The corresponding values of the distance are
$$b^2\text{ or }\frac12-b+(b-\frac12+b)^2.$$
You should be able to finalize the discussion.
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 $$ d = \begin{cases} \min(\lvert b \rvert, \sqrt{4b - 1}/2 & \text{for } b > 1/2 \\ \lvert b \rvert & \text{for } b \le 1/2 \end{cases} $$
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 $$ d = \begin{cases} \min(\lvert b \rvert, \sqrt{4b - 1}/2 & \text{for } b > 1/2 \\ \lvert b \rvert & \text{for } b \le 1/2 \end{cases} $$
Or, we can use a bit of geometry and say that the line of the shortest distance will be a normal from the point (0,b) to the parabola.
So, we take any arbitrary point (t,$t^2$) on the parabola and then find the slope of the normal at it,
Slope = -$dx/dy$ = -(1/2t).
Write the equation of the normal as
y-$t^2$=(-1/2t)(x-t).
This must pass through (0,b).
Satisfy the point in the equation to get,
t= $\sqrt {b-1/2}$ Now, the point on the parabola is- ($\sqrt{b-1/2}$, (b-1/2)). Use the distance formula to get the shortest distance as $\sqrt{b-1/4}$
