Finding tangent(s) of a cubic curve which also cut it orthogonally

Question

Tangents of a quadratic curve (circle and conics) cannot cut the curve again. However, for non-quadratic (cubic and other) curves it is possible. But it is interesting to ask if tangent(s) of a cubic curve for instance $y=3x^2-2x^3$ can cut it orthogonally at some other point.

The experience with this simple cubic has been interesting despite quadratic equations in two variables it is cumbersome to find such points on the curve and then the equation of tangent(s). We took the equations to Mathematica that gave four coordinates as involved irrational numbers.

The question is: Can there be either a simpler way of handling the ensuing equations or should we replace the present cubic by some other convenient cubic to get simpler equations and finally a model which is doable by hand? Also there me be some interesting generalization(s).

Answer 1

It's not so complicated!

You can assume the inflection point of the cubic at the origin. Up to similarity this means that we look at the cubic $$x\mapsto f(x):=x^3 - ax,\qquad a>0\ .$$ Let $\bigl(u,f(u)\bigr)$ be an arbitrary point on the cubic. The tangent there is then given by $$x\mapsto t(x)=f(u)+f'(u)(x-u)=(3u^2-a)x-2u^3\ .$$ One easily computes $$f(x)-t(x)=(u-x)^2(2u+x)\ .$$ This shows that the tangent intersects the cubic at the point $\bigl(-2u,f(-2u)\bigr)$. We therefore have to solve the equation $$f'(u)f'(-2u)+1=(3u^2-a)(12u^2-a)+1=0\ .$$ When $a>{4\over3}$ we have the four real solutions $$u=\pm\sqrt{{5a\pm\sqrt{9a^2-16}\over24}}\ .$$

See the Fig. below for $a=3/2$

Answer 2

Doing it with $y=3x^2-2x^3$

Let $T[t,3t^2-2t^3]$ and $N[u,3u^2-2u^3]$ so the slope of TN, slope of tangent at T and the slope of normal at N must be equal: $$3(t+u)-2(t^2+u^2+tu)=6t(1-t)=-\frac{1}{6u(1-u)}$$ The solution of the first and the second part; the first and the last part given by Mathematica is $$t=\left(\frac{1}{24}[12\pm\sqrt{15\pm\sqrt{17}}]\right).$$ It is a wonderful solution that remains to be obtained by hand!

Anyway the four possible roots give rise four possible tangents as $$y=6t(1-t)(x-t)+3t^2-2t^3$$ See the Fig. below which confirms the graphics of @almagest in the OP above. We must say that the cubic of @Christian Blatter in his answer above is the perfect model that is doable by hand.