Missing steps in proving 5 tangent lines determine a conic

Question

The question is simple: given lines $L_1,\ldots,L_5 \subset \Bbb{P}^2$, prove there exists a conic that is tangent to all 5 lines.

My solution goes along the lines of:

1. Every line $L_i$ in the projective plane corresponds with a point $\alpha_i$ in its dual
2. In the dual (which is also just the projective plane), we find a conic $\check{C}$ that passes through the 5 points
3. Expressing this dual conic in terms of the equation $x^T A x = 0$, we can define a conic $C$ in $\Bbb{P}^2$ represented by the matrix $\operatorname{Adj}(A)$

My question:

I now think I have to prove that the line $L_i$ is tangent to the conic $C$ if and only if the point $\alpha_i$ lies on the dual conic $\check{C}$. However, I don't know how to prove this. In particular, the "if" part is the more important one, but the "only if" part is annoyingly the easy implication. Can anyone help me further?

Some potential progress I have made:

• If we express the coordinates of $\Bbb{P}^2$ as $[x_0:x_1:x_2]$ and we say the line $L_i \leftrightarrow a_{i0}x_0+a_{i1}x_1+a_{i2}x_2 = 0$, then this line is tangent to $C$ at a point $p = [p_0:p_1:p_2]$ if and only if, as projective points, $$[a_{i0}:a_{i1}:a_{i2}] = \left[\sum_{k=0}^2 a_{0k}p_k:\sum_{k=0}^2 a_{1k}p_k:\sum_{k=0}^2 a_{2k}p_k\right]$$
• Obviously, the point $\alpha_i$ lies on $\check{C}$ if and only if $\alpha_i^T A \alpha_i = 0$

Answer 1

I will show you the way I solve it, by using a more geometric approach. In the end we will see that I basically do the same as you do, but the steps before will give me enough information to prove what I need.

Let $X=\begin{pmatrix}x_0\\x_1\\x_2\end{pmatrix}$. Then one can go to it's dual space by sending every line of the form $\begin{pmatrix}a_0&a_1&a_2\end{pmatrix}X=0$ to $\begin{pmatrix}a_0&a_1&a_2\end{pmatrix}$ and every point $\begin{pmatrix}a_0&a_1&a_2\end{pmatrix}$ to the line $\begin{pmatrix}a_0&a_1&a_2\end{pmatrix}X=0$.

The lines $L_i$ get mapped onto points $l_i$. Then find a conic $C$ through your the points $l_i$. Then find the tangents $T_i$ through the points $l_i$ to the conic $C$. Go back to your original projective space by doing the same transformations. Obviously the $5$ points $l_i$ get mapped onto the lines $L_i$ you started with. The lines $T_i$ get mapped onto $5$ points $t_i$ with $t_i$ a point of $L_i$. The conic through these $5$ points $t_i$ is the conic you are looking for.

Proof that this construction works

In the underlying proof, $i$ always ranges from $1$ to $5$.

Let $l_i$ be the vector such that the line $L_i$ has equation $l_i^TX=0$. This means the transformation to go to the dual space maps $L_i$ onto the point $l_i$. Let $X^TAX=0$ be the equation of the conic through all the $l_i$.

Then the tangents will be of the form $X^TAl_i$ so these tangents will correspond to the points $Al_i$ in the original space and these points obviously are on $L_i$.

Let $X^TBX=0$ be the conic through the points $Al_i$. Then $0=(Al_i)^TB(Al_i)=l_i^T(A^TBA)l_i$ for all $i$, so the matrix $A^TBA$ corresponds to the same conic as $A$ (since it is the unique conic through the points $l_i$) so they are the same up to some multiple $\lambda$. We get: $A^TBA=\lambda A\Longrightarrow B=\lambda(A^T)^{-1}=\lambda A^{-1}$ since $A$ is a symmetric matrix and where $\lambda$ depends on the choice of $B$ and choosing $B$ accordingly, one can let $\lambda=\det(A)$, which will imply that $B=\det(A)\cdot A^{-1}=\text{Adj}(A)$. So my construction is the same as yours.

Now we proof that the tangents at the points $Al_i$ are just the lines $L_i$. The tangents at $Al_i$ are $0=X^TB(Al_i)=X^T(\text{Adj}(A)\cdot A)l_i=\det(A)X^Tl_i$ where we can just divide by $\text{det}(A)$ to get the same equation as $L_i$. We see that $L_i$ are tangents for all $i$.

Extra notes:

I use in this proof without proving it that the tangent through the point $y$ to the conic with equation $X^TAX=0$ has equation $X^TAy=0$. This should be elementary knowledge. If you don't know it, I can spell out the details in a comment.

For more intuitive feeling about these things: The dual of a conic is the set of tangents to a conic. Using Steiner's definition for conics, this becomes pretty much trivial.