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?
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 $F=0$: $$(\gamma X)^2+(\gamma Y)^2+(-\alpha X-\beta Y)^2=\gamma^2X^2+\gamma^2Y^2+\alpha^2X^2+2\alpha\beta XY+\beta^2Y^2=$$ $$=(\alpha^2+\gamma^2)X^2+2\alpha\beta XY+(\beta^2+\gamma^2)Y^2=0.$$