Generators of conic pencil

Question

I’m studying linear algebra and conic sections.

CONTEXT AND TYPE OF EXERCISE

By conic pencil $F$ I mean the set of all conic sections that result from a linear combination of two conic sections $Γ_1, Γ_2$, that are said to generate $F$.

Conic pencils can be written in the form $$F: k_1Γ_1 + k_2Γ_2$$ where $k_1, k_2\in\mathbb{R}$ are real parameters and $Γ_1, Γ_2$ are two conic sections. So, conic pencils are generated by two conics.

Since the general equation for a conic is: $$Γ: ax^2 + by^2 + cxy + dx+ ey + f=0$$ we have a maximum of 6 parameters to determine a unique conic section, and we determine those parameters by using some conditions.

The type of exercises I’m interested in is the following:

1. You are given 5 indipendent conditions.
2. You use 4 conditions to choose the right conic sections $Γ_1$ and $Γ_2$ that generate a conic pencils $F$, so that you get an equation of type $F: k_1Γ_1 + k_2Γ_2$.

• Often they’re degenerate conic sections, since they’re simpler to see.

3. You say that one of the two generators, say $Γ_1$, doesn’t satisfy one of the 5 conditions: so we just delete the paramater $k_2$, because the fact that $Γ_1$ cannot be obtained for any value of $k_1$ doesn’t bother us (since $Γ_1$ isn’t a solution)
4. Thus, you have just one parameter left to determine the conic $Γ \in F$ that you’re searching (the equation is now of form $F: kΓ_1 + Γ_2$). You use the 5th condition to determine $k$ and you’re done.

• Note that point 3. and 4. aren’t really needed for question 1, but I think they help understand the kind of exercise.

QUESTION 1

My question is: how do we know that the chosen conic sections generate the conic bundle that we’re searching for, that satisfies the first 4 conditions, except by “brute forcing” all the possible generators and testing the 4 conditions?

Note that I want a method that is general, so that works for other conditions, not only points that are in a conic sections (for example, there might be “tangent to line” conditions, etc).

See examples below to understand my doubts.

EXAMPLE 1

Write the equation of a conic section Γ that satisfies the following: $$ P_1=(1, 0) \in Γ, P_2=(1, 1) \in Γ, P_3=(3, 0) \in Γ, P_4=(3, 3) \in Γ, P_5=(4,1) \in Γ$$

Solution (point 1.-2.): we choose $Γ_1: (x-1)(x-3)=0$, which satisfies $P_1, P_3 \in Γ$, and $Γ_2: y(x-y)=0$ which satisfies $P_2, P_4 \in Γ$.

My doubts: am I right, if I say that the only way to say that $Γ_1$ and $Γ_2$ actually generate the “right” conic pencil (and “right” in this context means “that includes the conic section that satisfies the 5th condition) is to see if the conic pencil satisfies the 5th condition? Are there any tricks to “already know” when you’re picking wrong generators of the pencil?

• In this case, we can see that the generators made by the solution satisfies all $P_i \in Γ$ because, if we plug in $P_i$ in $(x-1)(x-3)+k(y(x-y)=0$ it’s true.

For example, if I choose the conic sections $Γ_3: (y-x-1)(x-3)$ which satisfies $P_1, P_3 \in Γ$ and $Γ_4: y(x-y)$ which satisfies $P_2, P_4 \in Γ$, are they wrong just because I can see that the conic pencil $(y-x-1)(x-3) + ky(x-y)$ doesn’t actually satisfy $(1,1) \in Γ$? Because if I can’t see it before, it seems to me that I could get stuck in “trying a lot of different lines and never getting the right combination”. Am I missing something?

QUESTION 2

I don’t really understand point 3., it doesn’t really seem to make sense to me, but I found it in many solutions of this type of exercises. Can someone explain better why we really can just “avoid” one parameter, and why the outcome of that is really just excluding one of the generator and not excluding also other possible conic sections of the conic pencil that might be solutions?

EXAMPLE 2 (for question 2, asked by a comment)

Write the equation of a conic section $γ$ that satisfies the following: $$ P_1=(0, 0) \in γ, P_2=(1, 1) \in γ, P_3=(1, 2) \in γ, P_4=(2, 3) \in γ, P_5=\left(-\frac{1}{2},0\right) \in γ$$

We choose to build the conic pencil that satisfies every condition except $P_4 \in γ$. The first generator is the conic section $Γ_5: y(x-1) = 0$ which contains $P_1P_5$ and $P_2P_3$. The second generator is the conic section $Γ_6: (x-y)(4x-3y+2)=0$ which contains $P_1P_2$ and $P_3P_5$. The resulting conic pencil, for the solution, has equation $F: y(x-1) + t(x-y)(4x-3y+2)=0$ with $t\in\mathbb{R}$. As you can see, there is no parameter on $Γ_5$.

APPENDIX

See my other question.