# Exactly 2 tangents to every non-degenerate conic in the complex projective plane

Given a point $$p$$ not incident to a non-degenerate conic $$\mathcal{C}$$ in the complex projective plane $$\mathbb{C}\text{P}^2$$, how would you prove that there are exactly two tangents to $$\mathcal{C}$$ passing through $$p$$ ? I have found a proof in C. G. Gibson's book : Elementary geometry of algebraic curves, but I cannot seem to understand it.

I mainly do not understand the part where they substitute in $$F=0$$. Aren't the coordinates different?

• You should point out what you don't understand in that proof. – Aretino May 26 '19 at 21:18

From the first part you should understand that if a tangent goes through $$P=(\alpha : \beta : \gamma)$$ then it passes through the point $$(X:Y:Z)=(X:Y:\frac{-(\alpha X+\beta Y)}{\gamma})=(\gamma X, \gamma Y, -(\alpha X+\beta Y)$$ of the conic. Since this point is on the conic we can substitute its coordinates into $$F=0$$: $$(\gamma X)^2+(\gamma Y)^2+(-\alpha X-\beta Y)^2=\gamma^2X^2+\gamma^2Y^2+\alpha^2X^2+2\alpha\beta XY+\beta^2Y^2=$$ $$=(\alpha^2+\gamma^2)X^2+2\alpha\beta XY+(\beta^2+\gamma^2)Y^2=0.$$