Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Maybe someone knows this technique. Given $ F\left( {x,y} \right) \in {\Bbb C}\left[ {x,y} \right] $, let $ C = \left\{ {\left( {x,y} \right) \in {\Bbb C};F\left( {x,y} \right) = 0} \right\} $. Take a point $P \in C $ , consider the set L of all the lines that pass through $P$, clearly those lines are determined by their slope $\lambda$. Let now any $l \in L$ be any line through P. In some cases we can parameterize. Assume that for some reason there is a point of intersection of $l$ with $C$ , let´s call it $c(l)$ such that $c(l) = c(\lambda)= (c_x(\lambda),c_y(\lambda))$ Then clearly we have the parametrization, since $ c(\lambda) \in C $ so $ F(c(\lambda)) = 0 $

I have two questions about this, how can I prove that if there exist some point P, such that for any line that pass through P , there exist only one point (other than the fixed P), then I can write that intersection in terms of rational functions of the parameter $ \lambda $.

And also how can I prove using this technique the following problem? Let $ F(x,y) \in$ $ {\Bbb C}\left[ {x,y} \right] $ of degree 3 , such that $ F , F_x , F_y $ have a common zero , then there exist non constant polynomials $ f\left( t \right),g\left( t \right) \in {\Bbb C}\left( t \right) $ such that $ F(f(t),g(t))=0 $

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.