How many times tangent to a cubic curve $y = x^3$ from a point on it, meets again at another point.

Question

As per my elementary knowledge, the cubic curve can have a suitable start point so that if draw a tangent through it, it should intersect the curve at a point; then from this point of intersection can again draw a tangent that will intersect the curve. This means there are a total (maximum) of $3$ point linked together by tangents provided the first point on the curve is chosen suitably.

I want first of all vetting of the above, and the logic (reason) for the above statement.

I have tried manually and have no algebraic/ analytical reason to prove the above. But, my attempts yield only $3$ such points, 'provided' suitable start point is selected.

Answer 1

$$y=x^3$$

$$\frac{dy}{dx}=3x^2$$

Tangent at point $a$ is $$y-a^3=3a^2(x-a)$$

Let's figure out where else does it intersect.

$$x^3-a^3=3a^2(x-a)$$

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

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

$$(x-a)(x-a)(x+2a)=0$$

$$(x-a)^2(x+2a)=0$$

Hence the next point would be at $x=-2a$.

Hence if we started at $x=a$, the sequence will be a geometric sequence with common ratio $-2$, hence it can go up to infinitely many times.

Answer 2

This is an example of a wonderful phenomenon, encapsulated in Bézout’s Theorem, according to which two distinct irreducible curves, of degrees $d$ and $d'$ respectively, have precisely $dd'$ points of intersection.

Except that the points of intersection have to be counted with multiplicity; you have to look for them in an algebraically closed field (such as $\Bbb C$); and you have to look not in the ordinary “finite” plane, but in the projective plane.

In your case, you’re intersecting a curve of degree three (the cubic) with a curve of degree one (the line), in which case there will be precisely three intersection points, suitably counted. You’re drawing a tangent line to the cubic curve, that’s two points already (unless you draw it at the point $(0,0)$, where the tangent makes a triple contact), and the one remaining one has to have real coordinates: that is, you can see it on your graph.

Oh yeah, you say? What about two circles? Two curves of degree $2$ should have four points of intersection. Where are they? Well, I don’t want to go into the details of projective geometry, but there’s a pair points on the “line at infinity”, actually a pair of conjugate complex points, that every circle contains. And any two circles intersect at these points.