How do I prove that equation of pair of tangents to a conic is $T^2=SS_0$? Let $$\textbf M=\begin{bmatrix}a & h & g \\ h & b & f \\ g & f & c\end{bmatrix} \qquad \textbf r=\begin{bmatrix}x \\ y \\ 1\end{bmatrix}\qquad \textbf{r}_k=\begin{bmatrix}x_k \\ y_k \\ 1\end{bmatrix}\qquad \mathrm d\textbf{r}=\begin{bmatrix}\mathrm dx \\ \mathrm dy \\ 0\end{bmatrix}$$
Then equation of conic is $\mathrm S \equiv\textbf{r}^{\mathrm T}\textbf{Mr}=0$


*

*Firstly, $\ \vec{\textbf a}^{\ \mathrm T}\textbf{M}\vec{\textbf b}=\vec{\textbf b}^{\ \mathrm T}\textbf{M}\vec{\textbf a}$

*Next, $\mathrm {dS}=2\textbf{r}^{\mathrm T}\textbf{M}\mathrm d\textbf r$

*Equation of tangent at $\textbf r_1$ is $\textbf{r}_1^{\mathrm T}\textbf{M}\textbf r=0$

*Equation of Chord of contact from $\textbf r_0$ is $\textbf{r}_0^{\mathrm T}\textbf{M}\textbf r=0$

My attempts for proving $\mathrm{T^2=SS_0}$ is the pair of tangents.


*

*The pair of tangents is $(\textbf{r}_1^{\mathrm T}\textbf{M}\textbf r)(\textbf{r}_2^{\mathrm T}\textbf{M}\textbf r)=0$


*

*Sub Goal: to find a relation among $\textbf{r}_0$,$\textbf{r}_1$ and $\textbf{r}_2$


*The equation $(\textbf{r}_0^{\mathrm T}\textbf{M}\textbf r)^2-(\textbf{r}^{\mathrm T}\textbf{Mr})(\textbf{r}_0^{\mathrm T}\textbf{M}\textbf{r}_0)$ passes through $\textbf{r}_0$,$\textbf{r}_1$ and $\textbf{r}_2$. Thus if it is a degenerate hyperbola, then we are done.


*

*Sub Goal: $\det(\textbf{M}\textbf{r}_0\textbf{r}_0^{\mathrm T}\textbf{M}-(\textbf{r}_0^{\mathrm T}\textbf{M}\textbf{r}_0)\textbf{M})=0$


*Assuming pair of tangents can be written as $\mathrm{T^2-\lambda S - \mu T}=0$. Now equating slopes at 1 and 2 reveals $\mu =0$ and and since it passes through 0, we get $\lambda=\mathrm S_0$


*

*Sub Goal: proving the assumption.


 A: I will use another formulation so I apologize beforehand.
Considering
$$
M =\left( \begin{array}{cc}
a & b\\
b & c \end{array} \right), p = \left( \begin{array}{c}
x\\
y\end{array} \right), p_0 =  \left( \begin{array}{c}
x_0\\
y_0\end{array} \right), p_1 =  \left( \begin{array}{c}
x_1\\
y_1\end{array} \right), \vec v =  \left( \begin{array}{c}
v_x\\
v_y\end{array} \right)
$$
we have the conic $C$ and the line $L$
$$
\left\{
\begin{array}{rcl}
C & \rightarrow & (p_1-p_0)\cdot M \cdot(p_1-p_0) = c_0\\
L  & \rightarrow & p = p_1 + \lambda \vec v\\
\end{array}
\right.
$$
Now in $C \circ L = (p_1-p_0+\lambda \vec v) \cdot M\cdot (p_1-p_0+\lambda \vec v) = c_0$ tangency implies on
$$
\lambda^2\vec v \cdot M\cdot v +2\lambda (p_1-p_0)\cdot M\cdot \vec v +(p_1-p_0)\cdot M\cdot (p_1-p_0) = c_0
$$
$$
\Delta = (2(p_1-p_0)\cdot M\cdot \vec v))^2-4 (\vec v\cdot M\cdot\vec v)((p_1-p_0)\cdot M\cdot(p_1-p_0)-c_0) = 0
$$
From
$$
\lambda = \displaystyle{\frac{-2(p_1-p_0)\cdot M\cdot\vec v \pm \sqrt{\Delta}}{2\vec v\cdot M\cdot \vec v}}
$$
Now simplifying $\Delta$ we get at the tangency condition in terms of $\vec v$
$$
\vec v\cdot(c_0 M-\det(M)(p_1-p_0)\cdot N \cdot (p_1-p_0))\cdot \vec v = 0
$$
with $N = \left(\begin{array}{c,c}1& -1 \\-1&1\\\end{array}\right)$ and $\sqrt{v_x^2+v_y^2} = 1$
hence giving $C$ and $p_1$ we can choose $\vec v$ for tangency.
A: To show that the matrix $\mathbf M\mathbf r_0\mathbf r_0^T\mathbf M-(\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M$ is singular it suffices to show that it has a nontrivial null space. The null space of a line consists of the points on the line, so the null space of a pair of lines consists of the points on both lines. We therefore expect that $\mathbf r_0$ is a null vector of the matrix, and indeed $$\begin{align} [\mathbf M\mathbf r_0\mathbf r_0^T\mathbf M-(\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M]\mathbf r_0 &= \mathbf M\mathbf r_0\mathbf r_0^T\mathbf M\mathbf r_0-(\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M\mathbf r_0 \\ 
&= (\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M\mathbf r_0-(\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M\mathbf r_0 \\ &= 0\end{align}.$$  
This particular formulation of the tangent cone is related to the tangent-secant theorem. As well, it generalizes to higher dimensions: the tangent cone to a quadric $Q$ with vertex at $\mathbf v$ is $(Q\mathbf v)(Q\mathbf v)^T-(\mathbf v^TQ\mathbf v)Q$. (There’s another, and I think more straightforward, construction of the tangent cone in the plane, but it doesn’t generalize in the way this one does.)  
This expression looks a lot like an application of Plücker’s mu to me, and sure enough it can be derived using that method. $\mathbf M\mathbf r_0$ is the polar line of $\mathbf r_0$, and so the degenerate conic $(\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)^T = \mathbf M\mathbf r_0\mathbf r_0^T\mathbf M$ is a double line that passes through the intersection points of of the conic $\mathbf M$ and the tangents through $\mathbf r_0$. Applying Plücker’s mu, a conic that shares these intersection points and also passes through $\mathbf r_0$ is $$\begin{align} (\mathbf r_0^T\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)^T-\mathbf r_0^T(\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)^T\mathbf r_0\mathbf M &= (\mathbf r_0^T\mathbf M\mathbf r_0)[(\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)^T - (\mathbf M\mathbf r_0)^T\mathbf r_0\mathbf M] \\
&= (\mathbf r_0^T\mathbf M\mathbf r_0)[(\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)^T - (\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M]. \end{align}$$ If $\mathbf r_0$ does not lie on the conic, we can drop the factor of $\mathbf r_0^T\mathbf M\mathbf r_0$ and have the required expression. If on the other hand $\mathbf r_0$ lies on $\mathbf M$, the tangent at that point is its polar line so we have the double line $(\mathbf M\mathbf r_0)(\mathbf M\mathbf r_0)^T$ from which we can freely subtract the term $(\mathbf r_0^T\mathbf M\mathbf r_0)\mathbf M$ since in this case $\mathbf r_0^T\mathbf M\mathbf r_0=0$.
