0
$\begingroup$

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.enter image description here

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

$\endgroup$
  • $\begingroup$ You should point out what you don't understand in that proof. $\endgroup$ – Aretino May 26 '19 at 21:18
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Perfect, thank you. I had trouble figuring out which coordinates were getting substituted in where, but now I understand. $\endgroup$ – Shurik Goyal May 27 '19 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.