Lines tangent to the same conic Consider two lines $l_1, l_2\subset \mathbb{R}P^2$ and a projective transformation $\phi:l_1\to l_2$. How do I show that each line constructed by joining a point $P\in l_1$ with its image $\phi(P)$ is tangent to the same conic.
I tried letting $l_1= \lambda [(1,0,0)]+\mu[(0,1,0)]$, which is allowed after a projective transformation, but how to I determine the coefficients of the conic?
Any help would be very welcome!
 A: The fact that line $X\phi(X)$ envelopes a conic is a theorem due to Steiner (see here in particular the 7th figure).
As your question is about finding the equation of this conic, here is an answer giving this equation, and as well, proving the theorem.
I begin by an illustration:

Let us assume WLOG that the first line is taken as the axis of abscissa, which means that all its ordinary points have coordinates $(x,0,1)$ (its point at infinity is $(x,0,0)$).
WLOG again we can assume that the homography takes the form:
$$X=(x,0,1) \mapsto  \ \phi(X)=(ax+c,dx+f,gx+k)\approx\underbrace{(\frac{ax+c}{gx+k},\frac{dx+f}{gx+k},1)}_{\text{for "ordinary" points}}\tag{1}$$
for some constants $a,c,d,f,g,k$.
We are going to establish that the conic curve enveloped by line $X\phi(X)$ has the equation
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0\tag{2}$$
or said otherwise the matrix
$$\begin{pmatrix}A&B&D\\B&C&E\\D&E&F\end{pmatrix} \ \text{with} \ \begin{cases}A&=&d^2\\B&=&dk-ad-2fg\\C&=&(a-k)^2+4cg\\D&=&df\\E&=&af-fk-2cd\\F&=&f^2\end{cases}$$
Proof:
Due to (1), the equation of line $X\phi(X)$ can be written under the form of this determinant equal to 0:
$$\begin{vmatrix}x_0& (ax_0+c) &x&\\0&(dx_0+f)&y\\1&(gx_0+k)&1\end{vmatrix}=0$$
where $x_0$ is a parameter characterizing the straight line. Expanding:
$$x(dx_0+f)-y((ax_0+c)-x_0(gx_0+k))-x_0(dx_0+f)=0 \tag{3}$$
Now we are ready apply the standard method to find an envelope, i.e., differentiating (3) wrt parameter $x_0$ (see explanations in the appendix of one of my answers here):
$$dx-y(a-2gx_0-k)-2dx_0-f \implies x_0=\frac{f+ay-yk-dx}{2(gy-d)}\tag{4}$$
Plugging this expression of $x_0$ into (3), one obtains indeed the second degree equation (1).
Here is the Matlab program I have written for obtaining the figure:
    clear all;close all;hold on;axis([-4,4,-1,3])
    a=1;c=8;d=3;f=2;g=2;k=1
    for x=-4:0.2:4;
        plot([x,(a*x+c)/(g*x+k)],[0,(d*x+f)/(g*x+k)])
    end;
    A=d^2;
    B=-a*d+d*k-2*f*g;
    C=a^2-2*a*k+k^2+4*c*g;
    D=d*f;
    E=a*f-2*c*d-f*k;
    F=f^2;
    h=@(A,B,C,D,E,F,x,y)(A*x.^2+2*B*x.*y+C*y.^2+2*D*x+2*E*y+F);
    ez=ezplot(@(x,y)h(A,B,C,D,E,F,x,y))

