Conic pencils and conditions

Question

I have a problem in understanding the following solution to a problem regarding conic pencils.

EXERCISE

Let $F$ be the conic pencil in which every conic section $Γ$:

0.1. $P = (0, 1) \in Γ$

0.2. $Q = (0, 0) \in Γ$

0.3. $Γ$ is tangent to $r : x+y-1=0$ in $P$

0.4. $Γ$ is tangent to $s : y=0$ in $Q$

Determine the conic section $C_i \in F$ that satisfies the following condition $i$:

1. $C_1$ is a parabola
2. $R = (-1,1) \in C_2$
3. $S=(1,1) \in C_3$

SOLUTION

To generate $F$, let’s use two conics:

• The product of $r$ and $s$
• The line $t$ that satisfies $P, Q \in t$, counted two times

So we get the following equation: $tx^2 + y(x+y-1)=0$

The only problem with writing $F$ with “affine coordinates” is that we exclude the possibility to find $x^2=0$, but since it doesn’t satisfy any condition, it doesn’t matter.

And then the exercise continues by finding $C_1, C_2, C_3$…

MY QUESTIONS

Q.1 How does the solution generate F immediately? I wouldn’t know how to get an equation for F, except by making a system with 4 equations, each one for one of the conditions from 0.1 to 0.4. But since it seems that the solution is way faster, I wonder how to do it and if this way of generating conic pencils can be used for other exercises.

Q.2 What does “affine coordinates” mean in this context? Does it mean that we already applied an affine change of coordinates? Why does i timply that we exclude $x^2=0$

• Note that I’m not asking to solve points 1, 2 and 3 of the exercise, because I know how to do that.

Answer 1

I think I might have found the answer to my own question.

Conic bundles can be written in the form $kΓ_1 + hΓ_2$ where $k, h\in\mathbb{R}$ are real parameters and $Γ_1, Γ_2$ are two conic sections. So, conic bundles are generated by two conics.

Now, I can try to choose two conic sections that generate the conic bundle $F$, and then impose conditions 1 (or 2, or 3) to find a value of the parameter which gives me the conic section.

To try and find those two “generators”, I have to know the types of ocnic sections. Because saying “$F$ is tangent to $r$ in $P$“ is equal to saying that, in P, $F$ is exactly $r$, by applying the same reasoning for $s$ we find that a generator $Γ_1$ could be the product of $r$ and $s$, which is a type of conic section named “pair of distinct incident real lines”.

Then, we know that a line that satisfies the condition 0.1, 0.2 is $x=0$: if we count it two times, so $x^2=0$, we notice it’s a conic section named “double real line”, and it still satisfies the conditions.

Now, we have the equation of $F$: $$kx^2+hy(x+y-1)=0$$ We can verify that this equations satisfies 0.1, 0.2 (how do we verify that it also satisfies 0.3 and 0.4?) We can, additionally, delete the parameter $h$ of $y(x+y-1)$, because by doing that we only exclude the conic section $Γ_3: x^2=0$ (since there is no value of $k$ for which we can find $x^2=0$). Thus we have found the equation found by the solution of my question.

Answer 2

The conic bundle can be represented by the general equation

$A x^2 + B xy + C y^2 + D x + E y + F = 0 $

Applying the first two conditions, we get

$ C + E + F = 0 \hspace{15pt}(1) $

$ F = 0 \hspace{15pt}(2)$

Now compute the derivative $y'$ as follows

$ 2 A x + B (y + x y' ) +2 C y y' + D + E y' = 0 $

Now the line $x + y - 1 = 0 $ has a slope of $y' = -1$, therefore substituting $P(0, 1)$ into the above, one gets

$ B (1) + 2 C (1)(-1) + D - E = 0 $

i.e.

$ B - 2 C + D - E = 0 \hspace{15pt}(3)$

The line $y = 0 $ has a slope of $ y' = 0 $ , and by substituting $Q(0,0)$, we get

$ D = 0 \hspace{15pt}(4)$

Solving the system of linear equations $(1)-(4)$ yields,

$(A, B, C, D, E, F) = t (1, 0, 0, 0, 0, 0) + s (0, -1, -1, 0, 1, 0) $

Now for $C_1$, the only requirement is $ 4 A C - B^2 = 0 $

Hence,

$ 4 t (-s) - (-s)^2 = 0 $

Since $s \ne 0 $, then $ 4 t + s = 0 $

i.e. $ s = - 4 t $, thus

$ (A, B, C, D, E, F) = (t , 4 t , 4 t, 0, -4 t, 0 )$

Dividing by $t$, gives the required parabola

$ x^2 + 4 x y + 4 y^2 - 4 y = 0 $

For $C_2$ the point $R(-1, 1)$ is the on the conic, so

$t (-1)^2 - s (1)(-1) - s (1)^2 + s (1) = 0$

Hence $ t + s = 0 $, from which,

$ (A, B, C, D, E, F) = t (1, 1, 1, 0, -1, 0 ) $

Thus the equation of the conic $C_2 $ is

$ x^2 + x y + y^2 - y = 0 $

Finally, for $C_3$, the point $S(1, 1)$ is on the conic, so

$ t (1)^2 - s (1)(1) - s (1)^2 + s (1) = 0 $

Hence,

$ t - s = 0 $, from which

$ (A, B, C, D, E, F) = t (1, -1, -1, 0, 1, 0 ) $

And the equation of $C_3$ is

$ x^2 - x y - y^2 + y = 0 $