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!
Hello, Anemori!
I have a rather sloppy proof . . .
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 a quartic in $\displaystyle x$ . . . which has at most 4 roots.