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Let $L$ be a line in $\mathbb{P}^2$ and $\Gamma_L$ be the subspace of $\mathbb{P}^5$ parametrizing the conics of $\mathbb{P}^2$ that are tangent to or contain $L$. Prove that $\Gamma_L$ is a quadric hypersurface of rank 3.

Let $(x_0,x_1,x_2)$ be the coordinates of $\mathbb{P}^2$ and $(z_0,z_1,z_2,z_3,z_4,z_5)$ of $\mathbb{P}^5$. Now, taking the line to be $x_0=0$ it is easy to see that the surface in $\mathbb{P}^5$ corresponding to the conics that contain the line is the plane $z_3=z_4=z_5=0$. But how does one prove that the hypersurface corresponding to conics tangent to the line is a quadric of rank 3? Further, what is its relation to this plane?

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$\newcommand{\PP}{{\mathbb P}}$ Let (changing your notation) $X,Y,Z$ be the coordinates of $\PP^2$ and the general conic $C$ be given by

$$f(X,Y,Z) = a X^2+b X Y+ c Y^2+d X Z +e Y Z +f Z^2;$$

Now the condition that $C$ touches a general line $L$ tangentially can be computed with a special line $X + Y - Z = 0$ on an affine patch $Z = 1$.

So let

$$g(X,Y,Z) = X + Y - Z$$

On the affine patch with coordinates $x,y$ we have

$$F(x,y) = a x^2+b x y+c y^2+d x+e y+f;$$

and

$$G(x,y) = x + y - 1$$

Now set $x_1 = x_0 + \epsilon h_1$ and $y_1 = y_0 + \epsilon h_2$ with $\epsilon^2 = 0$ and substitute this into $F(x_1,y_1) = 0$ and $G(x_1,y_1) = 0$.

Taking coefficients of $1, \epsilon$ in $F(x_1,y_1)$ and $G(x_1,y_1)$ one gets four equations in $S=\{a,b,c,d,e,f,x_0,y_0,h_1,h_2\}$. Adding to these either $h_1 = 1$ or $h_2 = 1$ you have five equations from which you eliminate by groebner base calculation $T=\{x_0,y_0,h_1,h_2\}$. You get, going both ways, two polynomials $H_1(a,b,c,d,e,f) = H_2(a,b,c,d,e,f) = 0$.

These identical $H_1, H_2$ are the conditions on $a,b,c,d,e,f$ that $C$ is tangential to the line $X + Y - Z=0$.

The $H_i$ are

$$H_1 = H_2 = {e}^{2}-2\,ed+{d}^{2}-4\,fc-4\,dc-4\,fa-4\,ea-4\,ca+4\,fb+2\,eb+2\,db+ {b}^{2}$$

It is (with Maple) a trivial computation to get the $6 \times 6$ matrix corresponding to $H_i$ and compute its rank. It is indeed $3$.

The above proof is not wholly rigorous though, as there is no justification that the tangency condition to $X + Y - Z = 0$ is generic (besides from reproducing the result sought for). An alternative would be to try the same computation with $p X + q Y - r Z = 0$ and indeterminates $p,q,r$. I did this and got

$$H_1 = H_2 = {p}^{2}{e}^{2}+{q}^{2}{d}^{2}-4\,cf{p}^{2}-2\,dqpe+{r}^{2}{b}^{2}+4\,b fqp-4\,{q}^{2}af-4\,dprc+2\,berp+2\,dqrb-4\,{r}^{2}ac-4\,rqae$$

It is of bidegree $(2,2)$ in $a,b,c,d,e,f$ and $p,q,r$. The $6 \times 6$ matrix corresponding to $H_i$ now has quadratic polynomials in $p,q,r$ as entries. It is possible to compute its generic rank and one gets $3$ again.

Note, that the above calculation prescription with $x_1 = x_0 + h_1 \epsilon, y_1 = y_0 + h_2 \epsilon$ and the following groebner base calculation works for arbitrary $f(x_1,y_1) = 0 = g(x_1,y_1)$ and gives conditions of the existence of a common point $P$ with common tangent vector $v \in T_P g = T_P f$.

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