proof for that exactly one normal can be drawn to a parabola from an exterior point.

Question

It is well known that a maximum of three normals can be drawn from the interior of a parabola. But I am interested in the proof for the fact that exactly one normal can be drawn to a parabola from an exterior point wrt to the parabola.

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So far I have done this.

WLOG we consider $y^2=4ax$.

Let the exterior point be $(h,k)$.

We must have $$k^2-4ah> 0\tag1$$

Also the equation of a normal to a parabola is $$y=mx-2am-am^3$$ so $(h,k)$ satisfies $$am^3+2am-mh+k=0$$

All I have to do is show that there is only one real root possible for this cubic using $(1)$. Can anyone complete the proof?

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Alternative methods are welcome!

NoteBy a normal i mean that a segment that is external to parabola

Thank you!

Answer 1

From the cubic

$$ax^3 +(2a-h)x +k =0$$

and using the conventions in this paper, one has

$$\begin{align*} x_N &=0 \\ \\ \delta^2 &= \dfrac{h-2a}{3a} \\ \\ y_N &= k \\ \\ h' &= 2a\left(\dfrac{h-2a}{3a}\right)^\frac{3}{2}\\ \end{align*}$$

For simplicity and WLOG, assume $a >0$.

For the case of $h \le 2a$, then $\delta^2 \le 0$ and the cubic has no real turning points, and thus only 1 real root.

For the case of $h \gt 2a$, then $\delta^2 \gt 0$ and $h' \in \mathbb{R}$ and the cubic has two turning points. In this case, to have only 1 real root, one needs to satisfy

$$\left|\dfrac{-y_N}{h'}\right| = \left|\dfrac{-k}{2a\left(\dfrac{h-2a}{3a}\right)^\frac{3}{2}}\right| \gt 1$$

or equivalently

$$\begin{align*}\left|\dfrac{-k}{2a\left(\dfrac{h-2a}{3a}\right)^\frac{3}{2}}\right|^2 &\gt 1^2\\ \\ k^2 &\gt 4a^2\left(\dfrac{h-2a}{3a}\right)^3\\ \\ k^2 - 4a\left[a\left(\dfrac{h-2a}{3a}\right)^3\right] &\gt 0 \\ \end{align*}$$

Now, to be consistent with the constraint in (1), one needs to show that

$$\begin{align*}a\left(\dfrac{h-2a}{3a}\right)^3 &\le h \\ \\ \left(h-2a\right)^3&\le 27a^2h \\ \end{align*}$$

which is certainly not true for $h \gg 2a$.

Therefore, it appears the constraint in (1) is not sufficient to guarantee that there is only one real root for $h \gg 2a$.

Answer 2

The question claims that exactly one normal can be drawn to a parabola from a point outside the parabola. This is incorrect.

Presumably the OP is imposing the restriction that the segment of the normal joining the point to the parabola must itself be outside the parabola. In this case, here is a proof not involving cubics:

WLOG, assume $a > 0$. Like in the question, let $(h, k)$ be a point outside the parabola.

Let $(x_p, y_p)$ be a point on the parabola. Consider the normal to the parabola at $(x_p, y_p)$; let $(x_n, y_n)$ be a point on the normal. We can divide the normal into two “rays”: one has $x_n > x_p$ and is (at least partly) inside the parabola; the other has $x_n < x_p$ and is outside the parabola. We are only interested in the second. From now on, we refer to this “ray” as the normal.

We also want the possibility that $(h, k)$ is on the normal, so we now require $x_p > h$.

Let $k'$ be the value of $y_n$ when $x_n = h$. That is, let $(h, k')$ be a point on the normal. Also let $m = -\frac{1}{2a}y_p$ be the gradient of the normal. Then

$$k' = y_p + m(h - x_p) = y_p(1 + \frac{1}{2a}(x_p - h))$$

so

$$|k'| = |y_p|(1 + \frac{1}{2a}(x_p - h))$$

is a continuous, strictly increasing, unbounded function of $|y_p|$ that takes on the value $|k|$ for exactly one value of $|y_p|$.

Since $k'$ has the same sign as $y_p$, $k'$ is a one-to-one function of $y_p$ that takes on the value $k$ for exactly one value of $y_p$.

This shows that exactly one normal contains the point $(h, k)$, as required.

Answer 3

First of all, if by normal you mean any line that intersects the parabola at 90 degrees then this is not true: take any point with positive $x$ on $y=x^2$ and draw a normal line $l$ at $(x, x^2)$. It will intersect the part of the parabola with negative $x$s somewhere, say at $(x_n, x_n^2)$. Then any point to the left of that $(x_n, x_n^2)$ will have a normal which intersects $l$ at a point $p$ external to the parabola. This $p$ has two "normals" to the parabola. (This is why purely algebraic arguments that find such normal lines will not give uniqueness.) However, if by normal you mean a segment that is entirely external to the parabola, then this is true for the exterior of any closed convex set in $\mathbb{R}^n$ - hence for the exterior of the graph of any convex function (more precisely, for the "overgraph" $S=\{(\vec{x}, y)| y\geq f(\vec{x})\}$ for any convex function $f$), including of course the parabola $f(x)=ax^2+bx+c$.

Let's show this. Firstly, for any closed convex set $S$ and for any point $p$ not in $S$ there exists unique point $n$ in $S$ which is closest to $p$. This $n$ will lie on the boundary and, when the boundary is differentiable $np$ will be normal to it (see below for a proof of all of this). Moreover, since there are no points of $S$ closer to $p$ than $n$, the whole segment $np$ lies outside of $S$.

Now, why doesn't there exist a second normal segment? Suppose there is one, say $pn'$. Then the normal $T_{n'}$ to $pn'$ through $n'$ is tangent to the boundary of $S$ at $n'$. By convexity of $S$, $S$ lies on one side of this tangent $T_{n'}$. Since the segment $pn'$ is outside $S$, we must have $S$ on the opposite side of $T_{n'}$ from $p$. But then $n$ is on the opposite side of $T_{n'}$ from $p$, and in particular is further from $p$ than $n'$, which is a contradiction. This completes the proof.

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Now, for completeness, the existence of the closest point and it's normality:

Existence of the closest point: Pick a point $s\in S$ let $S_b=S\cap B(p, |p, s|)$ be the intersection of $S$ with closed ball around $p$. Now $S_b$ is closed and bounded, so the distance function to $p$ being continuous has a minimum on $S_b$. Suppose this minimum is $m$ and is achieved at $n\in S_b$. Now any other point in $S$ is either not in $S_b$, and so has distance to $p$ bigger than $|p-s|$ and so also bigger than $|p-n|$, or is in $S_b$ and so has distance to $p$ at least $|p-n|$. We conclude that $n$ is a point at which the minimum distance from $p$ to $S$ is achieved.

The closest point is on the boundary: Draw the segment $pn$. If $n$ is in the interior of $S$ there would be other points of the segment that lie in $S$; but then they would be closer to $p$ than $n$, which is impossible. So $n$ is on the boundary.

Uniqueness of the closest point: Draw the tangent $T_n$ to $S$ at $n$ (in general, "a support hyperplane"). Then all of $S$ is on one side of $T_n$ by convexity, and it is the opposite side from $p$ (otherwise some point $s' \in S$ is on the same side of $T_n$ as $p$ then, by convexity os $S$, the segment $s'n$ is in $S$, but for points on that segment close to $n$ the distance to $p$ is smaller than that of $n$, which is a contradiction). But all points on opposite side of $T_n$ from $p$ have higher distance to $p$ than $n$ -- and, in particular, all points of $S$ do. Thus $n$ is the unique point of $S$ closest to $p$.

Normality of the closest point: Now assuming the boundary $\partial S$ is differentiable, let's see why $pn$ is normal to $\partial S$. If $np$ is not normal to $\gamma$ at $n$, then the tangent $T_n$ of $\partial S$ at $n$ has points very near $n$ that are closer to $p$ than $n$; but then since near $n$ the $\partial S$ itself follows the tangent closely, there will be points of $\partial S \subset S$ that are closer to $p$ than $n$, which is not true. So $pn$ is orthogonal to $\partial S$ as wanted.