After some searching, I found the explanation in "Perspectives on Projective Geometry" by <PERSON>.
The problem is finding a line $g$ such that $g^TRg=0$, and $R=L^TQL$ where $Q$ describes the ellipse. The transformation with $L$ makes $R$ a dual conic consisting of two points of intersection and a double line that passes through them. $R$ is also a degenerate conic, so it has rank 1 or 2. For the rank 2 case, if R is symmetric (which I think it always is) it can be rewritten as $R = uv^T + vu^T$ (up to a scalar multiple), where $u, v$ are points where the line intersects the original conic. $uv^T$ alone can describe the degenerate conic since $$0 = \langle g,u \rangle \cdot \langle g, v \rangle = (g^Tu)(v^Tg)= g^T(uv^T)g$$ with $\langle g, u \rangle=0$ and $\langle g, v \rangle=0$ if $u$ and $v$ are on $g$, respectively. The product $uv^T$ is rank-1, so we need to find a matrix $W$ such that $g^T(R+W)g=0$ and $(R+W)$ is rank-1. 364
If $W$ is skew-symmetric, it will satisfy the first condition. The matrix sum then becomes: $$R+W=\left ( \begin{matrix} a&b&d\b&c&e\d&e&f\end{matrix} \right )+\left ( \begin{matrix}0&\tau&-\mu\-\tau&0&\lambda\\mu&-\lambda&0 \end{matrix} \right )$$ Notice that $M_w\,q=w \times q$ for any vector $q$ and the vector $w=[\lambda,\mu,\tau]^T$. We now need to solve for $w$ so that $(R+W)$ is rank-1. Notice that if the line $l$ passes through $u$ and $v$, $u \times v=\alpha l$, where $\alpha$ is a scalar we can choose. If we consider the sum $uv^T-vu^T$, we can see through expanding and analyzing the elements that: $$uv^T-vu^T = \alpha M_{u\times v}=M_{\alpha l}=\alpha M_l$$ Thus, $R+\alpha M_l = 2 uv^T$. Thus, we can then choose $\alpha$ to make $R+\alpha M_l$ rank-1. Since $L = M_l$, then we can say: $$L^TQL+\alpha L = uv^T$$ Since $u$ and $v$ are in homogeneous coordinates, any scalar multiples don't affect the points when we project them back into $\mathbb{R}^2$. 945