Consider the conic \(5 x^{2}-4 x y+2 y^{2}=30\) a. Write the left side in matrix form. b. Diagonalize the coefficient matrix, finding the eigenvalues and eigenvectors. c. Construct the rotation matrix from the information in part b. What is the angle of rotation needed to bring the conic into standard form? d. What is the conic?

Firstly, consider the conic equation \(5 x^{2}-4 x y+2 y^{2}=30\). We can write this equation in matrix form as follows:\[\begin{bmatrix}x\y\end{bmatrix}\begin{bmatrix}5 & -2\-2 & 2\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}= 30\]We have our coefficient matrix as \[A = \begin{bmatrix}5 & -2\-2 & 2\end{bmatrix}\]

To find the eigenvalues of the matrix A, it is needed to solve the characteristic equation, \(\text{det}(A - \lambda I) = 0\), where I is the identity matrix and \(\lambda\) represent the eigenvalues. We calculate the eigenvalue as follows:\(\text{det}(A - \lambda I) = \text{det}(\begin{bmatrix}5-\lambda & -2\-2 & 2-\lambda\end{bmatrix}) = 0\)This gives the equation \((5 - \lambda)(2 - \lambda) - 4 = 0\), which simplifies to \(\lambda^2 -7\lambda + 10 = 0\). Solving this quadratic equation, we find the roots \(\lambda = 2, 5\), so these are the eigenvalues of matrix A.Next, to find the eigenvectors, we solve the equation \(A v = \lambda v\)For each eigenvalue:For \(\lambda = 2\): We have\(\begin{bmatrix}5-2 & -2\-2 & 2-2\end{bmatrix}v = 0\)which simplifies to \(\begin{bmatrix}3 & -2\-2 & 0\end{bmatrix}v = 0\)Solve this system and one will have \(\mathbf{v}= (2, 3)\) as an eigenvector.Next, for \(\lambda = 5\):This gives\(\begin{bmatrix}5 - 5 & -2\-2 & 2 - 5\end{bmatrix}v = 0\)which simplifies to \(\begin{bmatrix}0 & -2\-2 & -3\end{bmatrix}v = 0\)Solve this system and one will have \(\mathbf{v}= (1, 0)\) as an eigenvector.

The rotation matrix can be constructed from the eigenvectors. Since each column of a rotation matrix are just the normalized eigenvectors of the matrix A, we have:\[R = \begin{bmatrix}2/\sqrt{13} & 1\3/\sqrt{13} & 0\end{bmatrix}\]The angle of rotation \(\theta\) can be calculated from the first column of the rotation matrix. Using the formula \(\theta = \arctan\left(\frac{y}{x}\right)\) (where \(x\) and \(y\) are the first and second elements of the first column in the above rotation matrix, respectively), we get \(\theta = \arctan(\frac{3/\sqrt{13}}{2/\sqrt{13}}) = \arccos(2/\sqrt{13})\) which can be simplified to \(\theta \approx 56.31^{\circ}\)

After rotating the conic by \(\theta\), we obtain the standard form of the conic. Since the eigenvalues are both positive and different from each other, the conic is an ellipse.