A conic $C$ has the equation: $x^2+4y^2-xy+x+2y=2$
Firstly how do I go about showing what type of conic it is by using the intersection at the point at infinity? I converted to homogeneous coordinates and set $z=0$ leaving me with $x^2+4y^2-xy=0$ I couldn't see a way of solving at all so had a guess it must be an ellipse. Was there a more eloquent way of doing this?
What is the intersection between the conic and the line $l=2y+x=1$
I have rearranged $l$ to find $x$ and substituted it into $C$ to solve for $y$.
$y(2y-1)=0, y=0$ or $y=1/2$
Subbed $y$'s back into $C$
When $y =0, x^2+x-2=0, x= 1$ or $x=2$
When $y=1/2, 2x^2+x=0, x=0$ or $x=-1/2$
I got: $(1,0),(-2,0),(0,1/2),(-1/2,1/2)$
However I feel this is wrong as if it is an ellipse it should intersect at most twice. What have I done wrong? Should I have subbed the values for $y$ into $l$ and not $C$?
Best Answer
Write down the conic's matrix (or matrices) (See here if you have doubts):
$$A=\begin{pmatrix} \;1&\!-\frac12&\frac12\\ \!-\frac12&4&1\\ \;\frac12&1&\!-2\end{pmatrix}\implies \det A=-8\neq0\implies\text{ the conic isn't degenerate}$$
We also have
$$\det\begin{pmatrix}1&\!-\frac12\\\!-\frac12&4\end{pmatrix}=\frac{15}4>0\;\implies\;\text{the conic is an ellipse}$$
About the intersection you've already been answered.