3.23. Derive the polar equation of the ellipse $x^{2} / a^{2}+$ $+y^{2} / b^{2}=1$ under the condition that the direction of the polar axis coincides with the positive direction of the x-axis, and the pole is at the center of the ellipse.
Solution. Substitute into the ellipse equation the expressions for $x$ and $y$ in terms of the polar radius $\rho$ and the angle $\varphi$.

We have

$$
\frac{\rho^{2} \cos ^{2} \varphi}{a^{2}}+\frac{\rho^{2} \sin ^{2} \varphi}{b^{2}}=1, \rho^{2} \frac{b^{2} \cos ^{2} \varphi+a^{2} \sin ^{2} \varphi}{a^{2} b^{2}}=1
$$

Replace $b^{2}$ in the numerator with $a^{2}-c^{2}$ and use the fundamental trigonometric identity

$$
\begin{aligned}
& \cos ^{2} \varphi+\sin ^{2} \varphi=1, \quad \rho^{2} \frac{\left(a^{2}-c^{2}\right) \cos ^{2} \varphi+a^{2} \sin ^{2} \varphi}{a^{2} b^{2}}=1 \\
& \rho^{2} \frac{a^{2}-c^{2} \cos ^{2} \varphi}{a^{2} b^{2}}=1, \frac{\rho^{2}}{b^{2}}\left(1-\frac{c^{2}}{a^{2}} \cos ^{2} \varphi\right)=1
\end{aligned}
$$

Substitute $\varepsilon^{2}=c^{2} / a^{2}$, then we get $\rho^{2}=b^{2} /\left(1-\varepsilon^{2} \cos ^{2} \varphi\right)$. 3.24. At what point does the linear function $L(x, y)=\sqrt{2} x-y$ reach its maximum in the region described by the inequality $x^{2} / 9+y^{2} / 7 \leqslant 1 ?$

Solution. Applying Theorems 3.3 and 3.4, we get that the maximum of $L(x, y)$ is achieved at the point of tangency of the right tangent to the given ellipse, parallel to the line $\sqrt{2} x-y=0$. Using condition 3.39:

$$
k=\sqrt{2}=\sqrt{2} x-2 y, a=3, b=\sqrt{7} ; 2 \cdot 9+7=m^{2}
$$

From the two values $m= \pm 5$ we choose $m=-5$, since the function $L(x, y)=\sqrt{2} x-y$ will reach its maximum in the given region in the direction of the vector $\vec{n}=\{\sqrt{2} ;-1\}$ on the right tangent $y=\sqrt{2} x-5$. Determine the coordinates of the point of tangency by substituting $y$ with $\sqrt{2} x-5$ in the ellipse equation. Hence

$$
\begin{aligned}
& \frac{x^{2}}{9}+\frac{(\sqrt{2} x-5)^{2}}{7}=1 \\
& 25 x^{2}-90 \sqrt{2} x+162=0, \Rightarrow x_{1}=x_{2}=1.8 \sqrt{2}
\end{aligned}
$$

Substitute $x_{1}$ into the tangent equation and find $y_{1}=$ $=-1.4$. For these $x_{1}$ and $y_{1}$, the linear function reaches the following maximum value:

$$
L_{\max }(1.8 \sqrt{2},-1.4)=\sqrt{2} \cdot 1.8 \cdot \sqrt{2}-(-1.4)=5
$$

## 3.5 . PROBLEMS

## STRAIGHT LINE