Minimum distance from a point on the axis of symmetry of a parabola to the parabola

Question

The equation of the parabola is $x^2=4ay$. The parametric equation is $x=2at$, $y=at^2$. I want to calculate minimum distance for any point on the axis of symmetry, i.e. $y$-axis here from the parabola curve.

For any point, $P(0,y)$ on the axis of symmetry, the square of the distance from the parabola is:

$$\begin{align} r^2 &= (2at)^2+(y-at^2)^2 \\ &=4a^2t^2+(y-at^2)^2 \end{align}$$

The differentiation w.r.t to $t$,

$$\frac{d(r^2)}{dt}= 8a^2t-4ayt+4a^2t^3$$

If we make, $d(r^2)/dt=0$ then,

$$\begin{align} t&=0 \tag{i} \\ t&=\pm \sqrt{\frac{y-2a}{a}} \tag{ii} \end{align}$$

Here we can see, the root from equation (ii) is imaginary until we take $y\geq 2a$. That means, for all the points starting from vertex up to $2a$ on the axis of symmetry, the minimum distance will be from vertex? Is it correct?

Answer 1

An easier way to solve this is as follows:

At a point $(0, p)$ on the parabola $y=c x^2$ the segment $(0, p)-(x, cx^2)$ is perpendicular to the slope of the parabola $(1, 2cx)$. So $$(1, 2cx).(x, cx^2-p)=0\Rightarrow 2c^3x^3=(2cp-1)x\Rightarrow x=0 \text{ or } x\pm \sqrt{\frac{2cp-1}{2c^2}}. $$ As in your notation $c=\frac{1}{4a}$, $p=y$ this simplifies to $t=\pm \sqrt{\frac{y-2a}{a}}$, so you have the correct answer.

Now to the question which point is closest. First note the two non-zero roots exists (i.e. are real, non-zero) iff $2c>1/p$, i.e. $y>2a$. Otherwise the minimum distance is just $y$ (as our distance function is bounded below, continuous etc.)

For $y>2a$, the distance to the points corresponding to $\sqrt{\frac{y-2a}{a}}$ can be brutally checked to be larger than for $t=0$. A more elegant argument for this is the following: $(0,0)$ can only be the closest point to $(0, y)$ if the circle with center $(0, y)$ and radius $y$ lies within the parabola. This can only occur if the curvature of the parabola at $(0, 0)$, i.e,. $1/2a$ is smaller than the radius of the circle $1/y$.

Thus $(0, 0)$ is the closest point iff $y\leq 2a$.

Answer 2

You can see below a purely geometrical solution: given point $A$ on the axis, $B$ is the nearest point only if line $AB$ is perpendicular to the parabola.

This is always the case if $B=V$, where $V$ is the vertex. But another solution is possible, when $AB$ is the bisector of the angle formed by $FB$ (where $F$ is the focus) with line $BH$ (parallel to the axis). It follows that $BH=AF$ (point $H$ on the directrix) and this solution then exists if $AF>FV$.

It is easy to check that this second solution, when it exists, gives the minimum distance. If the equation of the parabola is $x^2=4ay$ then $FV=a$ and $y_B=y_A-2a$.