Conics and tangents: a general method

   A conic or a $2^{nd}$ degree planar curve has the following general form in cartesian coordinates
$
\boxed{ax^{2}+by^{2}+cxy+dx+ey+f=0}
$
   If we denote with $C$ the graph of this equation, we are going to show that the equation of the tangent to the curve $C$ at the point $(x_{1},y_{1}) \in C$ can be immediately derived from the equation of the curve, using the substitutions:
$
\boxed{
\begin{array}{c}
x^{2} \rightsquigarrow xx_{1}, \  y^{2} \rightsquigarrow yy_{1} \\
 xy \rightsquigarrow \frac{1}{2}(xy_{1}+x_{1}y)  \\
x \rightsquigarrow \frac{x+x_{1}}{2}, \ y \rightsquigarrow \frac{y+y_{1}}{2} \\
\end{array}
}
$
Consequently, the equation of the tangent to the curve $C$ at the point $(x_{1},y_{1}) \in C$ can be immediately written as
\begin{equation} \label{tangentconic}
\boxed{
axx_{1}+byy_{1}+c\frac{xy_{1}+x_{1}y}{2}+d\frac{x+x_{1}}{2}+e\frac{y+y_{1}}{2}+f=0
}
\end{equation}
   Proof: $\bullet$ Since $(x_{1},y_{1}) \in C$ its coordinates should satisfy the equation of the curve, thus
\begin{equation} \label{pointonconic}
ax_{1}^{2}+by_{1}^{2}+cx_{1}y_{1}+dx_{1}+ey_{1}+f=0
\end{equation}
   $\bullet$ Now we can proceed in implicitly differentiating the initial equation with respect to $x$, obtaining thus an expression for its slope at the (arbitrary) point $(x_{1},y_{1}) \in C$:
$$
\begin{array}{c}
(ax^{2}+by^{2}+cxy+dx+ey+f)'=0 \Leftrightarrow \\
   \\
\Leftrightarrow 2ax+2byy'+cy+cxy'+d+ey'=0 \Leftrightarrow \\
    \\
\Leftrightarrow (2by+cx+e)y'=-2ax-cy-d \Rightarrow \\
    \\
\Leftrightarrow y'(x_{1})=\frac{dy}{dx}\mid_{(x_{1},y_{1})}=-\frac{2ax_{1}+cy_{1}+d}{2by_{1}+cx_{1}+e}
\end{array}
$$
for all $(x_{1},y_{1})$ for which  $2by_{1}+cx_{1}+e \neq 0$.
   $\bullet$ Consequently, the equation of the tangent to the curve $C$ at the point $(x_{1},y_{1}) \in C$ will be
$
y-y_{1}=\frac{dy}{dx}\mid_{(x_{1},y_{1})}\cdot(x-x_{1})
$
Thus we can now readily work out
$$
\begin{array}{c}
y-y_{1}=\frac{dy}{dx}\mid_{(x_{1},y_{1})}\cdot(x-x_{1}) \Leftrightarrow\\
    \\
\Leftrightarrow y-y_{1}=-\frac{2ax_{1}+cy_{1}+d}{2by_{1}+cx_{1}+e}(x-x_{1}) \Leftrightarrow \\
    \\
\Leftrightarrow (2by_{1}+cx_{1}+e)(y-y_{1})=-(2ax_{1}+cy_{1}+d)(x-x_{1}) \Leftrightarrow  \\
     \\
\Leftrightarrow 2by_{1}y-2by_{1}^{2}+cx_{1}y-cx_{1}y_{1}+ey-ey_{1}=-2ax_{1}x+2ax_{1}^{2}-cy_{1}x+cy_{1}x_{1} \\   -dx+dx_{1} \Leftrightarrow  \\
     \\
\Leftrightarrow 2by_{1}y+cx_{1}y+ey+2ax_{1}x+cy_{1}x+dx=2ax_{1}^{2}+cy_{1}x_{1}+dx_{1}+2by_{1}^{2}+ \\ cy_{1}x_{1}+ey_{1} \Leftrightarrow   \\
     \\
\Leftrightarrow 2ax_{1}x+2by_{1}y+c(x_{1}y+xy_{1})+dx+ey=  \\
  =\underbrace{ax_{1}^{2}+by_{1}^{2}+cx_{1}y_{1}}_{=-dx_{1}-ey_{1}-f}+\underbrace{ax_{1}^{2}+by_{1}^{2}+cx_{1}y_{1}+dx_{1}+ey_{1}}_{=-f} \Leftrightarrow   \\
    \\
\Leftrightarrow axx_{1}+byy_{1}+c\frac{xy_{1}+x_{1}y}{2}+d\frac{x+x_{1}}{2}+e\frac{y+y_{1}}{2}+f=0
\end{array}
$$
which finally completes the proof!