Example 3.38. Write the equation of the tangent plane and the normal line to the surface $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$ at the point $M\left(\frac{a}{\sqrt{2}} \cdot \frac{b}{2}, \frac{c}{2}\right)$.
Solution. The surface is given implicitly, so the corresponding equation of the tangent plane has the form (3.30). We find the partial derivatives involved in this equation, taking into account that in this case $F(x, y, z)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1$. We have: $F_{x}^{\prime}=\frac{2 x}{a^{2}} ; F_{y}^{\prime}=\frac{2 y}{b^{2}} ; F_{z}^{\prime}=\frac{2 z}{c^{2}}$. The values of the partial derivatives at point $M$ are:

$$
F_{x}^{\prime}(M)=\frac{\sqrt{2}}{a} ; F_{y}^{\prime}(M)=\frac{1}{b} ; F_{z}^{\prime}(M)=\frac{1}{c}
$$

Substituting the obtained results into (3.30), we get the desired equation of the tangent plane.

$$
\begin{gathered}
\frac{\sqrt{2}}{a}\left(x-\frac{a}{\sqrt{2}}\right)+\frac{1}{b}\left(x-\frac{b}{2}\right)+\frac{1}{c}\left(x-\frac{c}{2}\right)=0 \Rightarrow \\
\Rightarrow \frac{x \sqrt{2}}{a}+\frac{y}{b}+\frac{z}{c}=2
\end{gathered}
$$

To find the equation of the normal line, we substitute the values of the partial derivatives, calculated at the point, into (3.31). We get

$$
\begin{gathered}
\frac{x-\frac{a}{\sqrt{2}}}{\frac{\sqrt{2}}{a}}=\frac{y-\frac{b}{2}}{\frac{1}{b}}=\frac{z-\frac{c}{2}}{\frac{1}{c}} \Rightarrow \\
\Rightarrow \frac{a}{\sqrt{2}}\left(x-\frac{a}{\sqrt{2}}\right)=b\left(y-\frac{b}{2}\right)=c\left(z-\frac{c}{2}\right)
\end{gathered}
$$