Can you help me prove if true or false please:

True or False? Every system of two conic equations whose xy-term is nonexistent has at most 4 points of intersection. graph is not a proof.

Thanks!

Printable View

- May 11th 2010, 11:35 PMAnemoriConic Equation
Can you help me prove if true or false please:

True or False? Every system of two conic equations whose xy-term is nonexistent has at most 4 points of intersection. graph is not a proof.

Thanks! - May 12th 2010, 07:32 AMSoroban
Hello, Anemori!

I have a rather sloppy proof . . .

Quote:

True or False?

Every system of two conic equations whose xy-term is nonexistent

has at most 4 points of intersection. . . True!

We have two conic equations: . $\displaystyle \begin{array}{cccc} Ax^2 + By^2 + Cx + Dy + E &=&0 & [1] \\ Px^2 + Qy^2 + Rx + Sy + T &=& 0 & [2] \end{array}$

Solve [1] for $\displaystyle y\!:\quad By^2 + Dy + (Ax^2 + Cx + E) \:=\:0 $

. . Quadratic Formula: . $\displaystyle y \;=\;\frac{-D \pm\sqrt{D^2 - 4B(Ax^2 + Cx + E)}}{2B} $

Substitute into [2]:

. . $\displaystyle Px^2 + Q\left(\frac{-D \pm\sqrt{D^2 - 4B(Ax^2+Cx+E)}}{2B}\right)^2 + Rx \;+$ .$\displaystyle S\left(\frac{-D\pm\sqrt{D^2 - 4B(Ax^2+Cx+E)}}{2b}\right) + T \;\;=\;\;0$

This simplifies to ain $\displaystyle x$ . . . which has at most 4 roots.*quartic*