 ## Wednesday, May 14, 2014

### 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!