Intersection of perpendicular tangent lines - generalization of directrix?

Question

This is a funny little problem that I came up with.

For a differentiable function $f$, define a locus of points $P$ as follows:

• Let $m$ be an arbitrary tangent line to $f$, and let $n$ be another tangent line that is perpendicular to $m$. A point is in $P$ iff it is the intersection of $m,n$.

For a parabola, $P$ is the directrix (you can prove this with algebra or with geometry). But for other types of graphs, it gets a little weirder. Some examples:

• For $f(x)=x^3$, $P= \emptyset$ since $f'(x) \geq 0 \space \forall x$.
• For $f(x) = \sin(x)$, $P$ is a lattice of points consisting of the intersections of tangent lines having a slope of 1 and -1.

The specific problem I'm taking aim at is how to get an explicit description of what $P$ would look like for $f(x) = x^4$. I have drawn it in Geogebra and it looks like an inverted bell curve, which makes sense, but I can't get an actual equation to describe it. The same algebraic techniques used to show that $P$ will be the directrix when $f$ is a parabola end up with awful equations that I'm not even sure have a closed form solution.

The curve just looks so... I dunno... nice? I feel like there must be a much simpler answer than the pages of power functions that assail me every time I try this.

Answer 1

Very nice problem. As a novice algebraic geometer, this is my approach.

We start from the curve $C$ with equation $y=x^4$ on the plane $\pi$. If $(\bar x,\bar y)$ is a point of this curve, the line tangent to $C$ passing through this point is given by $(y-\bar y)=4\bar x^3(x-\bar x)$ which by using the equation $\bar y=\bar x^4$ can be rearranged into $y=(4\bar x^3)x+(-3\bar x^4)$.

Now, let's define a new space, called $L$. In this space, points represent what in the plane $\pi$ were lines: specifically, we represent the line in $\pi$ of equation $y=mx+q$ with the point $(m,q)$ in $L$ (these are not all the lines in $\pi$, but we don't care about those of the form $x=c$ as they are not tangent to $C$). The lines tangent to $C$ are thus represented by points of the form $(m,q)=(4\bar x^3,-3\bar x^4)$. Then $m^4=4^4\bar x^{12}$ and $q^3=-3^3\bar x^{12}$, so $3^3m^4+4^4q^3=0$: in other words, the set of lines tangent to $C$ are represented by a curve $C'$ in $L$, with equation $$3^3m^4+4^4q^3=0$$ (this is the so-called dual curve of $C$, in the dual space $L$).

We're interested in finding the points of intersection of tangent lines of $C$ which are perpendicular. Of course, two lines represented in $L$ by $(m,q),(m',q')$ are perpendicular if and only if $mm'=-1$; we can represent pairs of lines as points of the product space $L\times L$, and we call the subset of pairs of lines which are perpendicular $V$. With a bit of algebra it is possible to show that the intersection as lines in $\pi$ (if it exists) of $(m,q)$ and $(m',q')$ is the point $$\Big(\frac{q'-q}{m-m'},\frac{mq'-qm'}{m-m'}\Big).\tag{1}$$ We thus get what is called a birational map $\varphi:L\times L\to \pi$, which associates to pairs of lines their intersection (if it exists). The curve we're looking for is the image $\varphi(V\cap C'\times C')$, that is, intersections of pairs of tangent lines of $C$ which are perpendicular.

Let's find this image: $V\cap C'\times C'$ is given by equations $$\begin{cases} 3^3m^4+4^4q^3=0\\ 3^3m'^4+4^4q'^3=0\\ mm'=-1 \end{cases}$$ and by some rearrangements we get that $$q'^3=\frac{3^6}{4^8q^3}$$ and since we're dealing with real numbers, we can take the cube root of both sides to get $$q'=\frac{3^2}{4^{8/3}q}=\frac{\mu^2}{q}$$ where $\mu=3/(4^{4/3})$. Plugging this and $m'=-1/m$ in $(1)$ we get the point of $\pi$ $$(X,Y)=\Big(\frac{\mu^2m-q^2m}{m^2q+q}, \frac{\mu^2m^2+q^2}{m^2q+q}\Big).\tag{2}$$ Again, since we're dealing with real numbers, we can the cube root of both sides of the equation $3^3m^4=-4^4q^3$ to get, after some rearrangements, $$q=-\mu m^{4/3}=-\mu t^4$$ where $t=m^{1/3}$. Putting this in $(2)$ and after some algebra we finally get $$(X,Y)=\Big(\mu\frac{t^{11}-t^3}{t^{10}+t^4},-\mu\frac{t^8+t^6}{t^{10}+t^4}\Big)$$ which is a parametric equation representing the curve we're interested in!

Finally, we can get an implicit equation by following the procedure specified here. Using the software Sage with the following commands (I ignored the term $\mu$)

P.<x, y, t> = PolynomialRing(QQ, 3)
p = t^11 - t^3
q = -t^8 - t^6
r = t^10 + t^4

gcd = p.gcd(q).gcd(r)
pp = p.quo_rem(gcd)[0]
qq = q.quo_rem(gcd)[0]
rr = r.quo_rem(gcd)[0]

(x*rr-pp).resultant(y*rr-qq, t)

we get the equation $$x^4y^2 + 2x^2y^4 + y^6 - 2x^2y^3 - 2y^5 - 2x^2y^2 - y^4 + 2x^2y + 4y^3 - y^2 - 2y + 1=0$$ which (ignoring the scaling term $\mu$) looks correct! (You might need to put something like $0.99$ instead of $1$ to actually see something).