TITLE: Tangent line to a general conic at a point
QUESTION [5 upvotes]: If I have a conic
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
and want to know the tangent line at $(x_0, y_0)$, I thought I would just find the derivative (implicit) and use the equation of a line $(y-y_0) = m (x-x_0)$. For the derivative I get:
$$2Ax + By + \frac{dy}{dx}Bx + 2Cy\frac{dy}{dx} + D + E \frac{dy}{dx} =0$$
Solving for $\frac{dy}{dx}$ I get
$$m = \frac{dy}{dx} = \frac{-2Ax - By - D}{Bx + 2Cy + E}$$
Substituting this into the equation for a line:
$$(y-y_0) = \frac{-2Ax - By - D}{Bx + 2Cy + E} (x-x_0) = $$
and grouping the $x$, $y$, and constant terms, I get
$$(2Ax_o+By_0+D)x + (Bx_0+2Cy_0+E)y -2Bx_0y_0 - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0 = 0 $$
I've seen things like this: http://demonstrations.wolfram.com/TangentLinesToAConicSection/ that show an $F$ in the expression. The $x$ and $y$ terms are the same, but the constant term is $-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F$. In my method here, the $F$ term goes away in the differentiation immediately as it is a constant. Can anyone explain how these representations are equivalent?
$$-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F = - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0$$

REPLY [4 votes]: Theory: Equation of tangent at any point of the conic is given by $T = 0$.
$T$ is the transformed equation of the conic.
Following transformations are considered while transforming a conic at a point $(x_0, y_0)$
$$x^2 \implies x x_0$$
$$y^2 \implies y y_0$$
$$x \implies \frac{x+x_0}{2}$$
$$y \implies \frac{y+y_0}{2}$$
$$xy \implies x.y_0+y.x_0$$

Answer to your question: According to your question, the equation of conic is: 
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
Tangent at the point: $(x_0,y_0)$ will be given by $T$= $0$.
$\implies$ $T$=$0$
$\implies$ $$A(x.x_0) + B(x.y_0+y.x_0) + C(y.y_0) + D.\frac{(x+x_0)}{2} + E.\frac{(y+y_0)}{2} + F = 0$$
This is required equation of tangent.