Fast method to find the tangent line to a conic section: why does it work?

Suppose the original conic section is

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

The tangent line at $(x_0,y_0)$ is

$$y-y_0=m(x-x_0)$$

To find $m$, take derivative on both sides of the conic section:

$$2Ax+By+Bx\frac{dy}{dx}++2Cy\frac{dy}{dx}+D+E\frac{dy}{dx}=0$$ Solve for $\frac{dy}{dx}$ and plug into $m$:

$$y-y_0=\frac{-2Ax_0-By_0-D}{Bx_0+2Cy_0+E}(x-x_0)$$

Simplify:

$$(2Ax_0+By_0+D)x+(Bx_0+2Cy_0+E)y-(2Ax_0^2-2Bx_0y_0-2Cy_0^2-Dx_0-Ey_0)=0$$

Use the fact that $Ax_0^2 +Bx_0y_0 +Cy_0^2+Dx_0+Ey_0+F=0$ since $(x_0,y_0)$ is a point on the conic section, we can simplify the equation further into:

$$(2Ax_0+By_0+D)x+(Bx_0+2Cy_0+E)y+Dx_0+Ey_0+2F=0\Rightarrow\\ (Ax_0+\frac{1}{2}By_0+\frac{1}{2}D)x+(\frac{1}{2}Bx_0+Cy_0+\frac{1}{2}E)y+\frac{1}{2}Dx_0+\frac{1}{2}Ey_0+F=0$$

In your case, $B=E=0$:

$$(Ax_0+\frac{1}{2}D)x+(Cy_0)y+\frac{1}{2}Dx_0+F=0$$

that is

$$x\cdot x_0+y\cdot y_0+3x+3x_0-8=0$$

Here's a way to think of it. If you shift the point A (and the rest of the curve, with it's tangent line) to the origin you can look at the linear part at the origin (tangent line), then shift that line back to A. First shift A to (0,0):

$(x+x_A)^2+(y+y_A)^2+6(x+x_A)-8=x^2+2x\cdot x_A+x_A^2+y^2+2y\cdot y_A+y_A^2+6x+6x_A-8$

Now look at the linear part, and you get the tangent line (shifted to be through the origin). The linear part is $2x\cdot x_A+2y \cdot y_A +6x+6x_A-8+x_A^2+y_A^2$. Shift the origin back to A:

$2(x-x_A)x_A+2(y-y_A)y_A+6(x-x_A)+6x_A-8+x_A^2+y_A^2=2x\cdot x_A-2x_A^2+2y\cdot y_A-2y_A^2+6x-6x_A-8+x_A^2+y_A^2=2x\cdot x_A-x_A^2+2y\cdot y_A-y_A^2+6x-6x_A-8$.

Now use the fact that $x_A^2+y_A^2+6x_A-8=0$. You get the line $2x \cdot x_A+2y\cdot y_A +6x+6x_A-16=x \cdot x_A+y\cdot y_A +3x+3x_A-8=0$, which is the formula you gave. You can repeat this for an arbitrary quadratic, so any conic section.