Find a general formula for a tangent to a cubic which, is also normal to the cubic at the second intersection

Question

Let me preface by saying that I am not a professional mathematician, I just like to do some math in my spare time.

I assume someone has tried this before but I can't find the right words to type into a search engine to get me what I want.

As the title says, I would like some way to find the point on a general cubic such that a tangent at that point intersects the cubic once more at right angles.

I'd like to attempt it myself, and I've outlined a plan of attack:

• Let there exist a cubic polynomial $f(x)=ax^3+bx^2+cx+d$
• Set up the equation of the tangent at $x=x_T$. This would be $g(x)=f'(x_T)(x-x_T)+f(x_T)$
• Find the second intersection point of $g(x)$ and $f(x)$ in terms of $x_T$, call it $x_N$
• Let $f'(x_N)=-\frac{1}{f'(x_T)}$
• Solve for $x_T$ in terms of $a$, $b$, $c$, and $d$

Problems:

• Not all cubics would have this property. At the very least they should have two turning points, for which I know $b^2 > 3ac$, but I still don't think that's sufficient
• I have to solve a general cubic equation which I know has a formula but it's horrendous
• I don't even know if my plan of attack makes sense

Any help at all would be appreciated.

Note this is not a homework problem, just something I'm interested in solving.

Thank you.

Answer 1

No reason to consider any curve other than $y = x^3 - px,$ where $p > 0$ is constant.

Hard enough for $y = x^3 - 3 x.$ The point where we take the tangent is at $x=a,$ with $$ a = - \sqrt{ \frac{45 + \sqrt{585}}{72}} \; \; \approx \; \; -0.9802690478532443368005527022 $$

The tangent line at $(a, a^3 - 3a)$intersects the curve again when $x = -2a \; \; \approx \; \; 1.960538095706488673601105404$

The slope of the tangent line at $a$ is $-0.1172177814626812934541733462$

The slope of the tangent line at $-2a$ is $8.531128874149274826183306615 $

The product of these two slopes is $-1$ meaning the first tangent, at $a,$ is orthogonal to the curve at $-2a$

Need to check more, it is possible that $x=b,$ with $ b = - \sqrt{ \frac{45 - \sqrt{585}}{72}} \; \; \approx \; \; -0.5376547161709769327639370480 $ also works. I can see, this $b$ also works!

For your own pictures, the two tangent lines are $y$ set to $$ -0.1172177814626812934541733462 x + 1.883934787022560496398167095 $$

$$ -2.132782218537318706545826654 x + 0.3108424867671614744956344943 $$

Friday: if my $p$ is less than or equal to zero, a tangent to the graph of $y = x^3 - p x$ probably does intersect the curve again. However, a normal line to the curve never again intersects the curve. Below is a picture for $y=x^3$

Well, why not. Given real $\delta > 0,$ the slope of $y = x^3 + \delta x$ is $3x^2 + \delta$ and always positive. Any curve $y = g(x)$ that joins two distinct points of the cubic must have a point with positive slope, by the Mean Value Theorem. However, a normal line to the cubic has negative slope, therefore cannot intersect $y = x^3 + \delta x$ a second time.

Answer 2

As recommended by others, consider the cubic

$$y(x)=x^3-bx$$

The slope of the line joining $x_1,y(x_1)$ to $x_2,y(x_2)$ is

$$\frac{x_1^3-x_2^3-b(x_1-x_2)}{(x_1-x_2)}=x_1^2+x_1x_2+x_2^2-b$$

If this is also the slope of the tangent at $x_1$ we have $$x_1^2+x_1x_2+x_2^2-b=3x_1^2-b$$ which has the solution $x_2=x_1$ and the less trivial $x_2=-2x_1$. The condition that the tangents at $x_1,x_2$ are orthogonal is $$(3x_1^2-b)(3x_2^2-b)=-1$$ CORRECTION Inserting the previous result, we have $$(3x_1^2-b)(12x_1^2-b)=-1$$ $$\rightarrow 36x_1^4-15bx_1^2+b^2+1=0$$ $$\rightarrow x_1^2=\frac{5b\pm\sqrt{9b^2-16}}{24}$$

It can be easily shown that there are two real, positive, solutions for $x_1^2$ iff $b>4/3$. The illustration is for b=2.

Nice problem!

Answer 3

COMMENT.- A matter of principle must be stated: that the three real constants a, b, c are different, is equivalent to the cubic being elliptical: is of genus 1. When two or three of these constants are equal, the cubics are rational or of genus zero and there are three different classes of them: $$y^2=(x-a)(x-b)^2,\quad P=(b,0)\text { is a nodal point }.\\y^2=(x-a)^2(x-b,)\quad P=(a,0)\text { is a isolated point }.\\y^2=(x-a)^3,\quad P=(a,0)\text { is a cuspidal point }. $$ A curious fact is that for the curves with isolated point the plotters (Desmos and Wolfram, I don't know if other) do not put this point in their graphs.In the attached figure the black point $(1,0)$ is a point of both curves but it is not considered by the plotters.

What I want to add is that the more $b$ grows in relation to $a$, the cubic tends to be more and more "vertical" so it is a clear example that your problem has no solution for these kind of curves.

Answer 4

Here's an approach that starts at the point $(a, f(a))$ of intersection of degree $1$, rather than at the point $(x_T, f(x_T))$ at which the line is tangent.

Rescaling (changing coordinates by $(x, y) \rightsquigarrow (\tilde x, \tilde y) = (\lambda x, \lambda y)$, $\lambda > 0$), reflecting across the $x$-axis ($(x, y) \rightsquigarrow (-x, y)$) and translating ($(x, y) \rightsquigarrow (x - h, y + k)$ for constants $h, k$) preserve tangency and orthogonality of curves, so we may use the freedom of those coordinate changes to reduce the problem to the case $$f(x) = x^3 - C x .$$

Now, the line orthogonal to the graph of $f$ at $x = a$ is the graph of $$O(x) = f(a) - \frac{1}{f'(a)} (x - a) = a^3 - C a - \frac{x - a}{3 a^2 - C}$$ (except, if $C > 0$, where $a = \pm \sqrt\frac{C}{3}$, where the line is vertical, and hence does not intersect the graph of $f$ anywhere else).

Now, $f$ and $O$ have an intersection of multiplicity $1$ at $(a, f(a))$, so $O$ is tangent to $f$ if and only if $f - O$ has a multiple root. But a polynomial has a multiple root iff its discriminant $\Delta$ is zero. By construction, $f - O$ is a depressed cubic (that is, its quadratic coefficient is zero, so that it has the form $x \mapsto x^3 + p x + q$), so $$\Delta = -4 p^3 - 27 q^2 = \frac{[9 a^4 - 15 C a^2 + (4 C^2 + 4)][9 a^4 - 6 C a^2 + (C^2 + 1)]}{(-3 a^2 + C)^3} .$$

The second factor in the numerator is $(3 a^2 - C)^2 + 1 > 0$, so $\Delta = 0$ iff $$\phantom{(ast)} \qquad 9 a^4 - 15 C a^2 + (4 C^2 + 4) = 0 . \qquad (\ast)$$

(Solutions $(C, a)$.)

If $C < 0$, then the left-hand side is $(3 a^2)^2 + (\sqrt{- 15 C} a)^2 + 4 C^2 + 4 > 0,$ so there are no solutions---this recovers exactly the necessary condition $C \geq 0$ on the discriminant observed in the question statement (and specialized to our normal form for $f$).

Now, $(\ast)$ is a biquadratic equation in $a$, and completing the square in $a^2$, rearranging, and clearing denominators gives the equivalent equation $$(6 a^2 - 5 C)^2 = 9 C^2 - 16 .$$ We can now directly solve for $a$ as a function of $C$. In particular, we see that a necessary (and, one can see directly, sufficient) condition for the existence of a line satisfying the criteria is $C \geq \frac{4}{3}$. Moreover, we can see that:

• If $C > \frac{4}{3}$ there are four solutions, given by $$a = \pm \frac{1}{6} \sqrt{30 C \pm' 6 \sqrt{9 C^2 - 16}} .$$

(Typical generic case, $C = 2$)

• In the critical case $C = \frac{4}{3}$ there are two solutions, $$a = \pm \frac{\sqrt{10}}{3} .$$

(Critical case, $C = \frac{4}{3}$.)

Remark 1 Applying as appropriate a reflection, a rescaling, and a translation to bring the general form $f(x) = a x^3 + b x^2 + c x + d$ to our normal form $f(x) = x^3 - C x$, we see that the condition $C \geq \frac{4}{3}$ for the existence of solutions corresponds to the inequality $b^2 - 3 a c \geq 4 |a|$, which again is strictly stronger than the condition $b^2 - 3 a c \geq 0$ in the original question statement.

Remark 2 An interesting feature of this situation is that as $C \to \infty$, the horizontal spacing of the four orthogonal intercepts approaches $1 : 2 : 1$, and since this ratio is preserved by the applied transformations, the conclusion also applies to $f$ of the general form. Indeed, the asymptotic behaviors of the four solutions are $$\pm \sqrt{\frac{C}{3}}, \quad \pm 2 \sqrt{\frac{C}{3}} \pmod {O(C^{-3 / 2})} .$$