1.33. Reduce the quadratic form, which defines the equation of a second-order curve $5 x^{2}+6 x y+5 y^{2}-5 x-5 y=2$, to its canonical form.
Solution. The quadratic form defining the given equation is as follows:

$$
\begin{gathered}
F(x, y)=5 x^{2}+6 x y+5 y^{2}=5 x^{2}+3 x y+3 y x+5 y^{2}, \\
a_{11}=5, \quad a_{12}=a_{21}=3, \quad a_{22}=5
\end{gathered}
$$

The corresponding matrix $\tilde{A}$ of the quadratic form is:

$$
A=\left(\begin{array}{ll}
5 & 3 \\
3 & 5
\end{array}\right)
$$

We form the characteristic equation:

$$
\left|\begin{array}{cc}
5-\lambda & 3 \\
3 & 5-\lambda
\end{array}\right|=0 \Rightarrow(5-\lambda)^{2}-9=0 \Rightarrow \lambda_{1}=2, \lambda_{2}=8
$$

The quadratic form in the new coordinate system $O x^{\prime} y^{\prime}$ is written as follows: $F\left(x^{\prime}, y^{\prime}\right)=2\left(x^{\prime}\right)^{2}+8\left(y^{\prime}\right)^{2}$. We find the eigenvector for $\lambda_{1}=2$. We form the system:

$$
\left\{\begin{array} { l } 
{ ( 5 - \lambda _ { 1 } ) x ^ { \prime } + 3 y ^ { \prime } = 0 , } \\
{ 3 x ^ { \prime } + ( 5 - \lambda _ { 1 } ) y ^ { \prime } = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
3 x^{\prime}+3 y^{\prime}=0 \\
3 x^{\prime}+3 y^{\prime}=0
\end{array} \Rightarrow x^{\prime}+y^{\prime}=0 \Rightarrow y^{\prime}=-x^{\prime}\right.\right.
$$

Let $x^{\prime}=m$, where $m$ is any real number different from zero, the eigenvector corresponding to the eigenvalue $\lambda_{1}$ is

$$
x^{(1)}=m \vec{i}-m \vec{j}=\binom{m}{-m}
$$

We normalize this vector by dividing its coordinates by

$$
\left|x^{(1)}\right|=\sqrt{m^{2}+(-m)^{2}}=m \sqrt{2}: \tilde{x}^{(1)}=\frac{\sqrt{2}}{2} \vec{i}-\frac{\sqrt{2}}{2} \vec{j}=\binom{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} \text {. }
$$

the modulus,

We find the eigenvector for $\lambda_{2}=8$. We form the system:

$$
\left\{\begin{array} { l } 
{ ( 5 - \lambda _ { 2 } ) x ^ { \prime } + 3 y ^ { \prime } = 0 , } \\
{ 3 x ^ { \prime } + ( 5 - \lambda _ { 2 } ) y ^ { \prime } = 0 . }
\end{array} \Rightarrow \left\{\begin{array}{r}
-3 x^{\prime}+3 y^{\prime}=0, \\
3 x^{\prime}-3 y^{\prime}=0 .
\end{array} \Rightarrow x^{\prime}-y^{\prime}=0 \Rightarrow y^{\prime}=x^{\prime}\right.\right.
$$

Let $x^{\prime}=m$, where $m$ is any real number different from zero, the eigenvector corresponding to the eigenvalue $\lambda_{2}$ is

$$
x^{(2)}=m \vec{i}+m \vec{j}=\binom{m}{m}
$$

We normalize this vector by dividing its coordinates by the modulus,

$$
\left|x^{(2)}\right|=\sqrt{m^{2}+m^{2}}=m \sqrt{2}: \tilde{x}^{(2)}=\frac{\sqrt{2}}{2} \vec{i}+\frac{\sqrt{2}}{2} \vec{j}=\binom{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}
$$

We now write down the matrix of the linear transformation, the elements of which are the coordinates of the eigenvectors by rows:

$$
\tilde{B}=\left(\begin{array}{cc}
\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{array}\right)
$$

and the desired linear transformation is

$$
x=\frac{\sqrt{2}}{2} x^{\prime}-\frac{\sqrt{2}}{2} y^{\prime}, \quad y=\frac{\sqrt{2}}{2} x^{\prime}+\frac{\sqrt{2}}{2} y^{\prime}
$$

Substitute these dependencies into the original equation of the curve, taking into account that the quadratic form included in this equation has already been found:

$$
\begin{gathered}
2\left(x^{\prime}\right)^{2}+8\left(y^{\prime}\right)^{2}-5\left(\frac{\sqrt{2}}{2} x^{\prime}-\frac{\sqrt{2}}{2} y^{\prime}\right)-5\left(\frac{\sqrt{2}}{2} x^{\prime}+\frac{\sqrt{2}}{2} y^{\prime}\right)=2 \Rightarrow \\
\Rightarrow 2\left(x^{\prime}\right)^{2}+8\left(y^{\prime}\right)^{2}-5 \sqrt{2} x^{\prime}=2
\end{gathered}
$$

We complete the square in the obtained expression:

$$
\begin{gathered}
2\left(\left(x^{\prime}\right)^{2}-2 \cdot x^{\prime} \cdot \frac{1}{2} \cdot \frac{5 \sqrt{2}}{2}+\left(\frac{5 \sqrt{2}}{4}\right)^{2}\right)+8\left(y^{\prime}\right)^{2}=2 \Rightarrow \\
\Rightarrow\left(x^{\prime}-\frac{5 \sqrt{2}}{4}\right)^{2}+4\left(y^{\prime}\right)^{2}=1
\end{gathered}
$$

The coordinate transformation

$$
x^{\prime \prime}=x^{\prime}-\frac{5 \sqrt{2}}{4}, y^{\prime \prime}=y^{\prime}
$$

means a parallel shift of the axes to the new origin of coordinates

$$
\left(\frac{5 \sqrt{2}}{4} ; 0\right)
$$

Finally, we obtain the following canonical equation of the ellipse:

$$
\frac{\left(x^{\prime \prime}\right)^{2}}{1^{2}}+\frac{\left(y^{\prime \prime}\right)^{2}}{(0.5)^{2}}=1
$$