Finding intersections of two conics

[Scope of level: Undergrad Abstract Algebra]

This was a bonus topic we covered to very little extent so I'm rather unsure on how you go about doing this. From the notes we were given it considers the quadrics to be degenerates which then you can express them as factors of linear polynomials then solve separately.

If two quadratic form $P = Q = 0$ (of the form $ax^2 + 2bxy + cy^2 + 2dx +2ey + f$) are not degenerate then you construct an equivalent system that is degenerate by taking $(P') + k (Q')$ and find $k$ such that $P = P + k Q = 0$ where $(P')$ is a symmetric matrix constructed by taking the coefficients of the quadric $((P') = [(a,b,d),(b,c,e),(d, e, f)]$ by taking $det((P') + k(Q'))=0$.

(If need be, can provide link to said notes (in .pdf))

For example, consider the following two quadrics:

$2x^2 - xy + 3y^2 = 36$
$3x^2 - 4xy + 5y^2 = 36$

I tried the above steps and the constant required turns out to be some nasty number which really complicates computations (by quadratic formula as it forms a cubic polynomial on $k$ but with $(-36-36k)$ already factored out leaving a quadratic portion to be solved) which sends me the message that this is not right. So I just tried taking the difference of the two quadrics (which is essentially the case of $k=-1$) and solved directly as so and got $\{(4,2), (-4,-2), (3,3), (-3,-3)\}$ as my intersection points.

Is such approach acceptable or am I not seeing the picture here? Since I did not consider other possible values of $k$ I feel as if this misses out on some subset of solutions.

Any advice to point me in the right direction would be greatly appreciated :)

Solution 1:

Here is the rough picture. I say "rough" because if you want a precise result you have to add technicalities: projective plane instead of affine plane, intersection multiplicities, etc.

The one parameter family of conics given by $C_k=\lbrace P+kQ=0 \rbrace$ is called a pencil.
All the conics of this pencil go through the four intersection points $a,b,c,d$ of the two original conics $C_0=\lbrace P=0 \rbrace$ and $C_\infty=\lbrace Q=0 \rbrace$.
These four points are called the base points of the pencil and you have correctly identified them in your example.
Among the conics of the pencil there are three degenerate ones formed by a pair of lines.
Each pair is obtained by partitioning the base points $a,b,c,d$ in two pieces of two points and joining the points so obtained : for example one of the degenerate conics consists of the two lines $\overline {a,c}$ and $\overline {b,d}$.
The three degenerate conics can be computed by the procedure you describe of solving a cubic equation in $k$, but if you are so lucky as to know the base points, it is easier to just draw the lines between those lines and get the degenerate conics by the method of partitions described above.

The above considerations are part of a theme in algebraic geometry called linear systems.
A very elementary introduction to those can be found in chapter 16 of Gibson's Elementary Geometry of Algebraic Curves: An Undergraduate Introduction .