Why does the "T=0" method to calculate tangent work?

First of all, you probably know that the tangent at $x_0$ of the parabola $y=ax^2$ is: $$ y = ax_0^2 + 2ax_0(x-x_0). $$

If you didn't know this, keep reading! This means that, if we stay "close enough" to $x_0$, the line above is the one that approximates best the direction of the parabola.

In particular, if we move from $x_0$ to $x_0 + \Delta x$, we get: $$ a(x_0+\Delta x)^2 = ax_0^2 + 2ax_0\Delta x + a(\Delta x)^2. $$

You see that if $\Delta x$ is very small, the last term tends to vanish, and $$ a(x_0+\Delta x)^2 \approx ax_0^2 + 2ax_0\Delta x $$

becomes a very good approximation. Since $\Delta x = x-x_0$, the equation of the line given by: $$ y = ax_0^2 + 2ax_0(x-x_0) $$

is the equation of our desired tangent.

Your "T-formula" implies that a curve $C$ with equation: $$ ax^2 + 2hxy + by^2 + 2 gx +2 fy + c =0 $$

has tangent $T$ in $(x_0,y_0)$: $$ axx_0 + h(xy_0+x_0y) + byy_0 + g(x+x_0) + f(y+y_0) +c=0. $$

We would like to show now that $T$ is tangent, that is, that $T$ is the best possible approximation of $C$ if we stay "close enough" to the point.

If we move from $(x_0,y_0)$ to $(x_0+\Delta x, y_0+\Delta y)$, like before, we get: $$ a(x_0+\Delta x)^2 + 2h(x_0+\Delta x)(y_0+\Delta y) + b(y_0+\Delta y)^2 + 2 g(x_0+\Delta x) +2 f(y_0+\Delta y) + c =0. $$

Expanding the products: $$ \begin{array}{c} a(x_0^2+2x_0\Delta x+\Delta x^2)+ \\ + 2h(x_0y_0+\Delta xy_0+x_0\Delta y+\Delta x\Delta y)+ \\ + b(y_0^2+2y_0\Delta y+\Delta y^2)+ \\ + 2 g(x_0+\Delta x) +\\ +2 f(y_0+\Delta y) + c =0. \end{array} $$

Like before, we can neglect the quadratic "small" terms to get a linear equation, which approximates our curve in the best way possible. We obtain: $$ \begin{array}{c} a(x_0^2+2x_0\Delta x) + 2h(x_0y_0+\Delta xy_0+x_0\Delta y) + b(y_0^2+2y_0\Delta y)+ \\ + 2 g(x_0+\Delta x) +2 f(y_0+\Delta y) + c =0. \end{array} $$

To get the equation in $x,y$ for the line, we must (like before) replace $\Delta x$ with $(x-x_0)$, and now also $\Delta y$ with $(y-y_0)$. So: $$ \begin{array}{c} a(x_0^2+2x_0(x-x_0)) + 2h(x_0y_0+(x-x_0)y_0+x_0(y-y_0)) + b(y_0^2+2y_0(y-y_0))+ \\ + 2 g(x_0+x-x_0) +2 f(y_0+y-y_0) + c =0. \end{array} $$

Summing, we get: $$ \begin{array}{c} a(-x_0^2+2x_0x) + 2h(-x_0y_0+xy_0+x_0y) + b(-y_0^2+2y_0y)+ \\ + 2 gx +2 fy + c =0. \end{array} $$

This is an equation for the line, but we still haven't used the fact that the line must pass through our point $(x_0,y_0)$. Or, we still have to ensure that $(x_0,y_0)$ satisfies the equation for $C$. Now, we rewrite the expression above in the following way (we have moved the terms in $a,h,c$, and added and subtracted the terms in g,f,c): $$ \begin{array}{c} a(2x_0x) + 2h(xy_0+x_0y) + b(2y_0y) + g(x +x_0) + f(y+y_0) + 2c +\\ -(ax_0^2+2hx_0y_0+by_0^2+gx_0+fy_0+c) =0. \end{array} $$

Saying that $(x_0,y_0)$ passes through $C$ is saying that the second line vanishes! So we are finally left with: $$ 2ax_0x + 2h(xy_0+x_0y) + 2by_0y + g(x +x_0) + f(y+y_0) + 2c =0. $$

Dividing both sides by $2$, we get exactly your formula.

(Yes, it is a rather long and complicated process. I can't come up with a simpler one that doesn't lose clarity and doesn't use advanced calculus. This is probably why they never prove it in high school! Anyway, I'd be very very happy to see a shorter, elementary proof.)

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.