Find the conic tangent faster!

$\frac{d}{dx}(Ax^2+Bxy+Cy^2+Dx+Ey+F)=\frac{d}{dx}(0)$ $2Ax+B(x\frac{dy}{dx}+y)+2Cy\frac{dy}{dx}+D+E\frac{dy}{dx}=0$ $\frac{dy}{dx}=\frac{-(2Ax+By+D)}{Bx+2Cy+E}$ Put $(x_0, y_0)$ into $\frac{dy}{dx}$ , we get the slope of the tangent, and hence the equation of the tangent is $\frac{y-y_0}{x-x_0}=\frac{-(2Ax_0+By_0+D)}{Bx_0+2Cy_0+E}$ $(y-y_0)(Bx_0+2Cy_0+E)=-(x-x_0)(2Ax_0+By_0+D)$ $Bx_0y+2Cy_0y+Ey-Bx_0y_0-2Cy_0^2-Ey_0=2Ax_0^2+Bx_0y_0+Dx_0-2Ax_0x-By_0x-Dx$ $0=2Ax_0^2+2Cy_0^2-2Ax_0x+B(2x_0y_0-x_0y-y_0x)-2Cy_0y+D(x_0-x)+E(y_0-y)$ $0=Ax_0^2+Cy_0^2-Ax_0x+B(\frac{2x_0y_0-x_0y-y_0x}{2})-Cy_0y+D(\frac{x_0-x}{2})+E(\frac{y_0-y}{2})$ Since $(x_0, y_0)$ lies on the conic, so $Ax_0^2+Bx_0y_0+Cy_0^2+Dx_0+Ey_0+F=0$ , or equivalently, $Ax_0^2+Cy_0^2=-(Bx_0y_0+Dx_0+Ey_0+F)$

Hence the tangent becomes $-(Bx_0y_0+Dx_0+Ey_0+F)-Ax_0x+B(\frac{2x_0y_0-x_0y-y_0x}{2})-Cy_0y+D(\frac{x_0-x}{2})+E(\frac{y_0-y}{2})=0$ $-Ax_0x+B(\frac{-x_0y-y_0x}{2})-Cy_0y+D(\frac{-x-x_0}{2})+E(\frac{-y-y_0}{2})-F=0$ Dividing both side by $-1$ , we finally get our equation of tangent $\boxed{Ax_0x+B(\frac{x_0y+y_0x}{2})+Cy_0y+D(\frac{x+x_0}{2})+E(\frac{y+y_0}{2})+F=0}$

Result: If we want to find the equation of tangent of a conic at a point (on the conic), we just have to change the term $x^2$ , $y^2$ , $xy$ , $x$ , and $y$ systematically.