# Proof that if a tangent line has two intersections with a conic, that the conic is degenerate

I'm reading the book "Multiple View Geometry in Computer Vision", and there is a proof in there that I need help understanding. The following is all in $$\mathbb{P}^2$$ space and in homogeneous representation.

The theorem in question is:

The line $$l$$ tangent to a conic $$C$$ at a point $$x$$ on $$C$$ is given by $$l = Cx$$.

I can understand the proof for the non-degenerate case:

Given the line $$l = Cx$$, we know it passes through $$x$$ since $$l^Tx = x^TCx = 0$$. This comes from the fact that $$x$$ is a point that lies on $$C$$. If $$x$$ is the only intersection, the conic is non-degenerate and we're done.

However, for the degenerate case, the proof goes on to say that, suppose $$l$$ also intersects the conic at another point $$y$$, then $$y^TCy = 0$$, and $$x^TCy = 0$$ since $$x^TC = l^T$$ and $$l^Ty = 0$$. From this it follows that $$(x + \alpha y)^TC(x + \alpha y) = 0$$ for all $$\alpha$$, which means that the whole line $$l = Cx$$ joining $$x$$ and $$y$$ lies on the conic $$C$$.

The part in bold is verbatim and I can't figure out how that conclusion was made. Could someone explain how $$x^TCy \iff (x + \alpha y)^TC(x + \alpha y) = 0$$, and how to visualize such a degenerate case on a 2D projection?

From $$x^TCy=0$$ it also follows $$y^TCx=0$$ and $$(x+\alpha y)^TC(x+\alpha y)= x^TCx+\alpha x^TCy+\alpha y^TCx+\alpha^2 y^TCy=0.$$