Let $M$ be the matrix of the conic $$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$ so we have: $$M=\left[\begin{array}{ccc} A&\displaystyle{\frac{B}{2}}&\displaystyle{\frac{D}{2}}\\ \displaystyle{\frac{B}{2}}&C&\displaystyle{\frac{E}{2}}\\ \displaystyle{\frac{D}{2}}&\displaystyle{\frac{E}{2}}&F\end{array}\right]$$ What is the simplest way with no Differential Calculus to show that:
The coefficients $P_1,Q_1,R_1$ equation of the Polar Line $P_1x+Q_1y+R_1=0$ that passes by an exterior point (a pole) $(x_0,y_0)$ is given by the matrix product of $M$ and the column ${\mathbf{x}}=(x_0,y_0,1)^{t}$, that is: $$ M\left[ \begin{array}{c} x_0\\ y_0\\ 1 \end{array} \right]=\left[ \begin{array}{c} P_1\\ Q_1\\ R_1 \end{array} \right]$$
I tried considering a line that passes through $(x_0,y_0)$: $$P_1(x-x_0)+Q_1(y-y_0)=0$$ and considering two separated cases: $P_1\neq 0$ and $Q_1\neq 0$ to replace $x$ or $y$ in the conic equation and saying that the discriminant of the resulting 2nd degree equation is zero to get conditions on $P_1$, $Q_1$ and $R_1$ to show the assertion but I got stuck.
this image came from http://mathworld.wolfram.com/Polar.html
Consider the map $\pi$ from the set of points to the set of lines given by matrix product with $M$. You want to check that $\pi(P)=AB$.
Step 1. Check the following proposition: if $\pi(X)$ passes through $Y$, then $\pi(Y)$ passes through $X$. Use that $M$ is symmetric. Note: a line $Px+Qy+R=0$ passes through a point $(x_0,y_0)$ if and only if for $p=(P,Q,R),x=(x_0,y_0,1)$ we have $x\cdot p^t=0$ (here $t$ denotes the transpose of a matrix).
Step 2. Show that it is enough to check that $\pi(A)$ is the tangent $AP$ (similarly, $\pi(B)=BP$).
Step 3. Check that $\pi(A)$ passes through $A$ and it is the only point of intersection of $\pi(A)$ and the conic. Use that a line $XY$ has parametric equation $tX+(1-t)Y$.