Why does the "T=0" method to calculate tangent work? Given a random equation of a curve: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Suppose we need to find the tangent to this curve at any point $A(x_1, y_1)$. A method given to me by my professor was the 'T' method:

The 'T' form of an equation can be obtained by replacing:
  $$x^2 \rightarrow xx_1$$
  $$y^2 \rightarrow yy_1$$
  $$x \rightarrow \frac{x + x_1}{2}$$
  $$y \rightarrow \frac{y + y_1}{2}$$
  $$xy \rightarrow \frac{xy_1 + x_1y}{2}$$

The tangent to the curve is then the equation $T=0$.
For instance, if we need to find the tangent at $(2, 2)$ to the parabola $y^2 - 2x=0$:
$T =0$: $\implies yy_1 - 2\frac{x+x_1}{2} = 0$
Substituting $x_1 = 2$ and $y_1 = 2$: $$2y - (x+2) = 0$$ $$\implies 2y - x = 2$$
which is the required tangent.
I don't understand how this works! Could someone help me understand why it does?

This also works for other cases like:


*

*Deriving the equation of the two tangents to a curve from a certain external point: $SS_1 = T^2$ (S is the equation of the curve and S1 is the value given by the equation when the point is substituted into it (the power of the point wrt the curve))

 A: 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: 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.)
