# Polar curve of a non singular projective cubic curve with respect to inflection point is union of two distict lines.

Hey guys I am currently struggling with a question that goes as follows.

Let $$C$$ be a non-singular projective cubic and let $$p \in C$$ be an inflection with tangent line $$T$$. Show that the polar curve of $$C$$ with respect to $$p$$ is formed by two distinct lines $$T$$ and $$L$$ and that $$L$$ intersects $$C$$ in $$3$$ different points two by two.

Okay, let's say the curve $$C$$ is determined by the equation $$f$$ and say $$p=[a_p:b_p:c_p]$$ so the polar curve $$P$$ is defined up to multiplication by a constant by the polynomial $$f_P = a_p\frac{\partial f}{\partial x} + b_p \frac{\partial f}{\partial y} + c_p\frac{\partial f}{\partial z}$$. I also know that the intersection multiplicity $$I_p(T^*, C)$$ in $$p$$ of any tangent line $$T^*$$ through $$p$$ must be greater or equal to three because of the fact that $$p$$ is an inflection point. Now by Bezout's Theorem it is clear that there can only be one tangent line through $$p$$.

I'm not sure how to finish this. For one I'm not even sure how to prove that the tangent line is part of the polar, thereafter how to prove that there must be another line shouldn't be too difficult as the obtained formula $$f_P$$ is a degree 2 polynomial. Proving that the line $$L$$ is never tangent to $$C$$ also seems difficult from where I stand at the moment.

Any help will be greatly appreciated, just a little nudge in the right direction will probably suffice.

Thanks!

Let the cubic be given by $$f=y^2z−(x^3+pxz^2+qz^3)=0,$$ where $$x^3+px+q$$ doesn't have a double root. All non-singular cubic plane curves can be moved to this weierstrass form. Also let $$(\alpha:\beta:\gamma)$$ be a flex. The conic $$\alpha\frac{\partial f}{\partial x} + \beta \frac{\partial f}{\partial y} + \gamma\frac{\partial f}{\partial z}=0$$ has matrix $$\begin{pmatrix}3\alpha&0&p\gamma\\0&-\gamma&-\beta\\p\gamma&-\beta&p\alpha+3q\gamma\end{pmatrix}$$ which has determinant a multiple of the Hessian of $$f,$$ which vanishes on flexes. So the polar curve is a degenerate conic.

Indeed in the affine $$\gamma=1,z=1$$ the center is $$(−\frac{p}{3\alpha},−\beta)$$ and the conic is a pair of lines $$3\alpha(3x+p)^2−(3\alpha(y+\beta))^2+3\alpha(3\alpha\beta^2+3p\alpha^2+9q\alpha−p^2)=0$$ because the Hessian $$3\alpha\beta^2+3p\alpha^2+9q\alpha−p^2$$ vanishes. Explicitly the lines are $$\pm\sqrt{3\alpha}(3\alpha x+p)=3\alpha(y+\beta).$$

The flex $$(\alpha,\beta)$$ is on the line pair since $$12\alpha\beta^2-(3\alpha^2+p)^2=(3 \alpha \beta^2+3 p \alpha^2+9 q \alpha-p^2) -9\alpha(-\beta^2+\alpha^3+\alpha p+q).$$

Also, the tangent line at the flex $$(-3\alpha^2-p)(x-a)+2\beta(y-\beta)=0,$$ is in the ideal

$$\langle y^2-x^3-px-q,\\\beta^2-\alpha^3-p\alpha-q,\\3\alpha\beta^2+3p\alpha^2+9q\alpha-p^2,\\3 \alpha (3 \alpha x+p)^2-(3 \alpha (y+\beta))^2\rangle\\ +\langle x-\alpha,y-\beta\rangle^2$$

where we've intersected by the double of the flex.

However, which one of $$\pm\sqrt{3 \alpha} (3 \alpha x+p)=3 \alpha (y+\beta)$$ is the tangent line at $$(\alpha,\beta),$$ depends on $$p,q.$$

To the question, $$T$$ takes three of the six contacts we have by Bezout, so $$L$$ must also meet the curve in three with multiplicity. However $$T\cap L$$ is outside the curve, and moreover if $$L$$ were to be tangent, it couldn't be at a flex. We're left with three distinct intersection points of $$L$$ with the curve.