While geodude's method works, I'm putting up my calculus version of the problem.
Taking the random curve: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Differentiating it wrt x:
$$2ax + 2hx\frac{dy}{dx} + 2hy + 2by\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$$
$$\implies (ax + hy + g) + \frac{dy}{dx}(hx + by + f) = 0$$
$$\implies \frac{dy}{dx} = -\frac{(ax + hy + g)}{(hx + by + f)}$$
The equation of the tangent at the point $(x_1, y_1)$ is: $(y-y_1) = \frac{dy}{dx} (x - x_1)$
$$\implies (y-y_1)(hx+by+f) = (x_1 - x)(ax + hy + g)$$
$$\implies hxy + by^2 + fy -hxy_1 - byy_1 - fy_1 = axx_1 + hyx_1 + gx_1 - ax^2 - hxy - gx$$
Rearranging:
$$ax^2 + by^2 + 2hxy + gx + fy = axx_1 + hyx_1 + gx_1 + hxy_1 + byy_1 + fy_1$$
Adding $gx + fy + c$ on both sides:
$$ax^2 + by^2 + 2hxy + 2gx + 2fy +c = axx_1 + hyx_1 + gx_1 + hxy_1 + byy_1 + fy_1 +gx + fy + c$$
The LHS is $0$.
$$\therefore axx_1 + 2h\bigg(\frac{x_1y + xy_1}{2}\bigg) + byy_1 + 2g\bigg(\frac{x_1 + x}{2}\bigg) + 2f\bigg(\frac{y + y_1}{2}\bigg) + c = 0$$
Which is the "T" form of the equation.
A more general proof:
Let Q and R be the points at which lines through $P=(x_1,y_1)$ touch a non degenerate conic $S(x,y) \equiv Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$. In other words, lines PR and PQ are the tangents to this conic at points Q and R, and RQ is the chord of contact.
Let $PR(x,y)=0$, $PQ(x,y)=0$, $RQ(x,y)=0$ be the equation of these lines.
As RQ is polar of P in relation to this conic,
$$RQ(x,y)\equiv (Ax+By+D)x_1+(Bx+Cy+E)y_1+(Dx+Ey+F)=0$$
On the other hand, the equation $\lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$ represents all conics which are touched by lines PR and PQ at points R and Q. Therefore, for especific values of $\lambda$ and $\mu $ (none of which can be equal to zero, because otherwise S would be a degenerate conic):
$$S(x,y)\equiv \lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$$
Then,
$$S(x_1,y_1)=\lambda(PR(x_1,y_1).PQ(x_1,y_1))+\mu(RQ(x_1,y_1))^2,$$
$$S(x_1,y_1)=\mu(RQ(x_1,y_1))^2$$
Besides that,
$$RQ(x_1,y_1)=(Ax_1+By_1+D)x_1+(Bx_1+Cy_1+E)y_1+(Dx_1+Ey_1+F),$$
$$RQ(x_1,y_1)=S(x_1,y_1)$$
Thus
$$S(x_1,y_1)=\mu(S(x_1,y_1))^2,$$
$$\mu=\frac {1}{S(x_1,y_1)}$$
Therefore
$$S(x_1,y_1).S(x,y)\equiv S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))+(RQ(x,y))^2,$$
$$S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))\equiv S(x_1,y_1).S(x,y)-(RQ(x,y))^2$$
Finally, equating left and right members of this identity to zero, we get that the equation of tangents PR and PQ to conic S can be represented by equation
$$S(x_1,y_1).S(x,y)-(RQ(x,y))^2=0$$
Best Answer
The linear equations $L=0$ and $S_1=0$ are equivalent iff the coefficients are proportional. So, you should use the equations $$\frac{g+x_1}{l}=\frac{f+y_1}{m}=\frac{gx_1+fy_1+c}{n},$$ and by solving the linear system $$\begin{cases} \displaystyle\frac{g+x_1}{l}=\frac{f+y_1}{m}\\ \displaystyle\frac{gx_1+fy_1+c}{n}=\frac{f+y_1}{m} \end{cases}$$ that is $$\begin{cases} mx_1-ly_1=lf-mg\\ mgx_1+(mf-n)y_1=nf-mc \end{cases}$$ we find $$x_1 = \frac{lf^2-mgf+ng-lc}{lg+mf-n}\quad,\quad y_1 = \frac{mg^2-lfg+nf-mc}{lg+mf-n}.$$