802. Write the equation of the tangent and normal to the curve $x^{2}-2 x y+3 y^{2}-2 y-16=0$ at the point $(1 ; 3)$.
Solution. Substituting the coordinates of the given point into the equation, we verify that the point $(1 ; 3)$ lies on the curve

$$
1-2 \cdot 1 \cdot 3+3 \cdot 3^{2}-2 \cdot 3-16=0 \text { or } 0=0
$$

Differentiating with respect to $x$, we find

$$
2 x-2 y-2 x y^{\prime}+6 y y^{\prime}-2 y^{\prime}=0
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-164.jpg?height=409&width=411&top_left_y=357&top_left_x=77)

Fig. 62 from which

$$
y^{\prime}=\frac{x-y}{x-3 y+1}
$$

The derivative at the point $(1 ; 3)$

$$
y^{\prime}(1)=\frac{2}{7}
$$

The equation of the tangent line:

$$
y-3=\frac{2}{7}(x-1)
$$

or

$$
2 x-7 y+19=0
$$

The equation of the normal line:

$$
y-3=-\frac{7}{2}(x-1) \text { or } 7 x+2 y-13=0
$$