TITLE: Why does the "T=0" method to calculate tangent work?
QUESTION [9 upvotes]: 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))

REPLY [8 votes]: 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.