Example 2. Derive the equations of the tangent and normal to the curve $x^{2}+3 y^{2}=4$, drawn at the point with ordinate $y=1$ and negative abscissa.
Solution. When $y=1$, we find $x=-1$ and $x=1$. The problem mentions the point $A(-1,1)$. We find $y^{\prime}$ from the equation $x^{2}+3 y^{2}=4$. We have $2 x+6 y \cdot y^{\prime}=0$, hence $y^{\prime}=-\frac{x}{3 y}$, and thus, substituting $x=-1$, $y=1$, we get $k=y^{\prime}=\frac{1}{3}$.

Therefore, $(t): y=1+\frac{1}{3}(x+1),(n): y=1-3(x+1)$.

Answer. $(t): y=\frac{1}{3} x+\frac{4}{3} ; \quad(n): y=-3 x-2$.