Determining a conic and points of intersection between it and a line

Question

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 $C$?

Answer 1

I'm not sure about the first half of your question, but I will answer the second, about the points of intersection with the line

---

We have \begin{align}x^2+4y^2-xy+x+2y&=2\tag{1}\\ 2y+x&=1\tag{2}\end{align}

Rearranging $(2)$ gives us:

$$x=1-2y$$

We can substitute this into $(1)$ to give us:

\begin{align}(1-2y)^2+4y^2-(1-2y)y+(1-2y)+2y&=2\\ 4y^2-4y+1+4y^2-y+2y^2+1-2y+2y&=2\\ 10y^2-5y+2&=2\\ 10y^2-5y&=0\\ 5y(2y-1)&=0\\ y(2y-1)&=0\end{align}

So we have $y=0$ or $y=\cfrac 12$

When $y=0$, we can substitute into equation $(2)$ to see that$$x=1-2\times 0=1$$

When $y=\cfrac 12$, we can substitute into equation $(2)$ to see that$$x=1-2\times \frac 12=0$$

So we have points of intersection at $(1,0)$ and $(0,\frac 12)$

---

If we instead substitute $y=0$ into equation $(1)$, then we get

\begin{align}x^2+4\times 0^2-x\times 0+x+2\times 0&=2\\ x^2+0-0+x+0&=2\\ x^2+x-2&=0\\ (x+2)(x-1)&=0\end{align}

And so we have $x=-2$ and $x=1$.

We then check these answers in equation $(2)$ and see that $$2\times 0+-2 = -2\neq 1$$ but $$2\times 0+1=1$$ and therefore $x=1$ is the only valid solution to the equation.

Basically, the $x^2$ term in equation $(1)$ adds invalid solutions, so we must always check answers by substituting in the values of $x$ and $y$, to ensure we haven't introduced extra answers. It is also good practice (and easier for you!) to substitute into the simpler equation of the two to start with - in this case, equation $(2)$.

Answer 2

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.

Answer 3

You need to check the intersection of the conic with the line at infinity to determine its nature, not with “the” point at infinity, of which there are an infinite number. That looks like what you did, though, leading to the equation $$x^2-xy+4y^2=0$$ which has no real solutions, so you indeed have an ellipse. Another somewhat roundabout way to accomplish this is to compute the polar point of the line at infinity and plug it into the ellipse equation. We have the matrix $$Q=\left[\begin{array}{rrr}1&-\frac12&\frac12\\-\frac12&4&1\\\frac12&1&-2\end{array}\right]$$ for this conic, so the polar point of the line at infinity is $$(0:0:1)\,Q^{-1}=\left(\frac14:\frac18:-\frac28\right)=(2:1:-2).$$ This calculation isn’t as involved as it might seem at first glance. The left-hand side is just the last column of $Q^{-1}$, so it can be computed by augmenting $Q$ with $(0,0,1)^T$ and row-reducing. Plugging this point into the homogeneous equation of the conic yields $-24$, which means that the polar point is inside the conic and so the line at infinity doesn’t have any real intersections with it. Of course, once you’ve constructed the matrix $Q$, you can instead compute its determinant and principal minor to determine the nature of the conic as DonAntonio explains in his answer.

As far as your extraneous solutions go, they came from substituting the values of $y$ that you’d computed for the intersections back into the equation of the ellipse. For each of these values of $y$, there’s a pair of points on the ellipse with that ordinate, but only one point of each pair is on the intersecting line. You would’ve been better off substituting back into the equation of the line instead, which would’ve given you only one solution for each $y$-value. Barring that, when you’re dealing with squares (or higher powers) of variables in systems of equations, there’s the possibility of extraneous solutions being introduced by the squaring, so you should always check that the results you get satisfy all of the equations in the system.