Line tangent to conic section Let $l_1$ and $l_2$ be two lines in the real projective plane $\mathbb{R}P^2$ and let $\phi: l_1 \rightarrow l_2 $ be a projective transformation. Show that every line constructed by connecting a point $P \in l_1$ and its image $\phi(P)$ is tangent to the same conic section.
I tried this exercise as follows: Let $Q$ be the intersection of the two lines $l_1$ and $l_2$. We can assume (after a projective transformation) that $Q = E_0 = [(1,0,0)]$. Since we have to come op with a conic section, I thought that the asked conic section was then the cone with top $E_0$, call it $V(f)$. The function $f$ is then equal to
$$ f(x_1,x_2) = a x_1^2 + b x_2^2 + c x_1 x_2, \quad a,b,c \in \mathbb{R}, $$
since $V(f)$ is a cone with top $E_0$ and therefore $f$ is independent of $x_0$.
Now I am kinda stuck; I do not know how to proceed. I would like to give conditions to the variables $a,b$ and $c$ but I do not see how.
Thanks for helping me!
 A: Let $A,B\in l_1$ be two points on the first line, and let $C=\phi(A)$ and $D=\phi(B)$ be the corresponding points on the second line. I want to assume
$$\phi(\lambda A+\mu B)=\lambda C+\mu D\quad$$
which works fine as long as we use the same representatives (same scalar multiples) for $A$, $B$ and the transformation matrix $\phi$ in all the computations. Switching representative to a different member of the same equivalence class would break this correspondence.
Now a generic member of your bundle of lines can be described as
$$(\lambda A+\mu B)\times(\lambda C+\mu D)=\lambda^2(A\times C)+\lambda\mu(A\times D+B\times C)+\mu^2(B\times D)$$
with $\lambda,\mu\in\mathbb R;(\lambda,\mu)\neq(0,0)$. Each of those lines is tangent to the conic you want to find.
This looks familiar to me. Are you familiar with the rational parametrization of the unit circle? The points
$$\begin{pmatrix}\lambda^2-\mu^2\\2\lambda\mu\\\lambda^2+\mu^2\end{pmatrix}=
\lambda^2\begin{pmatrix}1\\0\\1\end{pmatrix}+
\lambda\mu\begin{pmatrix}0\\2\\0\end{pmatrix}+
\mu^2\begin{pmatrix}-1\\0\\1\end{pmatrix}$$
all lie on the unit circle
$$(x,y,z)\cdot\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}
\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}=0$$
And because the unit circle is self-dual (the matrix is its own inverse), you could as well make the same statement about lines of that form being tangent to the unit circle.
So at this stage, you should have an intuitive understanding that we are looking at a projectively transformed version of the unit circle, and therefore a generic conic section.
I'll leave the details to you to work out. You might want to actually formulate the projective transformation matrix which maps those cross product expressions to the tangents of the unit circle, and then use that matrix to transform the unit circle and obtain your desired conic. You also might want to discuss the case when that transformation is not invertible, and when the conic you get from your line correspondence would be degenerate.
I also happen to remember that Richter-Gebert's Perspectives on Projective Geometry covers this topic as well. Theorem 10.2 in section 10.3 is essentially your statement, but with additional non-degeneracy conditions: the two initial lines must be distinct, and the bundle of connecting lines must not all pass through a single point. Skimming his proof, I see it is based more strongly on cross ratios, which are central to his approach to conic sections.
