1.25. Find the equation of the tangent to the conic $a x^{2}+$ $+b x y+c y^{2}+d x+e y+f=0$ at the point $\left(x_{0}, y_{0}\right)$.
1.25. Suppose that the point $\left(x_{0}, y_{0}\right)$ belongs to the conic

$$
a x^{2}+b x y+c y^{2}+d x+e y+f=0
$$

Any line passing through the point $(x_{0}, y_{0})$ is given by the equation $y-y_{0}=k\left(x-x_{0}\right)$ (or the equation $x=x_{0}$, which corresponds to $k=\infty$). Let's find the second point of intersection of the line and the conic. For this, we substitute the equation of the line into the equation of the conic (i.e., replace $y$ with $\left.k\left(x-x_{0}\right)+y_{0}\right)$ and compute the coefficients $A$ and $B$ of $x^{2}$ and $x$ (the constant term $C$ will not interest us):

$$
\begin{gathered}
A=a+b k+c k^{2} \\
B=-b k x_{0}+b y_{0}-2 c k^{2} x_{0}+2 c k y_{0}+d+k e
\end{gathered}
$$

The considered line is tangent to the conic if and only if it intersects the conic only at the point $\left(x_{0}, y_{0}\right)$. An equivalent condition is: $2 x_{0}=-B / A$, i.e.,

$$
k\left(b x_{0}+2 c y_{0}+e\right)=-\left(2 a x_{0}+b y_{0}+d\right)
$$

Thus, the equation of the tangent line is

$$
\left(x-x_{0}\right)\left(2 a x_{0}+b y_{0}+d\right)+\left(y-y_{0}\right)\left(b x_{0}+2 c y_{0}+e\right)=0 .
$$

By the condition, the point $\left(x_{0}, y_{0}\right)$ belongs to the conic, i.e., $a x_{0}^{2}+b x_{0} y_{0}+$ $+c y_{0}^{2}+d x_{0}+e y_{0}+f=0$. Using this equality, the equation of the tangent line can be written in another form:

$$
\left(2 a x_{0}+b y_{0}+d\right) x+\left(b x_{0}+2 c y_{0}+e\right) y+d x_{0}+e y_{0}+2 f=0
$$