Find the equations of the following conics and quadric surfaces relative to principal axes

$\mathbf{8}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{8}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{35}$

Quadratic equation $8{\mathrm{x}}^2+8\mathrm{xy}+2{\mathrm{y}}^2=35$.

A scalar associated with a given linear transformation of a vector space and having the property that there is some nonzero vector which when multiplied by the scalar is equal to the vector obtained by the transformation operate on the vector especially: a root of the characteristic equation of a matrix.

For the following quadratic equation given by

$8{\mathrm{x}}^2+8\mathrm{xy}+2{\mathrm{y}}^2=35$ ……….. (1)

Which represent a central conic section with centre at the origin $A{x}^2+2Hxy+B{y}^2=K$where $A,B,H$ and K are constants.

To use the diagonalization process, transform the following quadratic equation given in equation (1) in a matrix form defined by

$\left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}A& H\\ H& B\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=Kor\left(\begin{array}{cc}x& y\end{array}\right)M\left(\begin{array}{c}x\\ y\end{array}\right)=K$……… (2)

Here M represent the $2\times 2$square matrix defined by

$M=\left(\begin{array}{cc}A& H\\ H& B\end{array}\right)$

Therefore, the following quadratic equation given in equation (1) can be written in a matrix form using equation (2) which is

$\left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}8& 4\\ 4& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=35$

Here $M=\left(\begin{array}{cc}A& H\\ H& B\end{array}\right)=\left(\begin{array}{cc}8& 4\\ 4& 2\end{array}\right)\mathrm{and}K=35$

For the following $2\times 2$ square matrix M defined by

$M=\left(\begin{array}{cc}8& 4\\ 4& 2\end{array}\right)$

To find its eigenvalues, find the values of $\lambda$ which satisfy the characteristic equation of the matrix M, thus the values of $\lambda$must satisfy $\begin{array}{c}\begin{array}{c}det\left(M-\lambda I\right)=\left|M-\lambda I\right|\\ =0\end{array}\end{array}$here I is the $2\times 2$square identity matrix.

Computing the matrix$M-\lambda I$implies that

$$\begin{gathered} M-\lambda I=\left(\begin{array}{cc}8& 4\\ 4& 2\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}8& 4\\ 4& 2\end{array}\right)-\left(\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right) \\ =\left(\begin{array}{cc}8-\lambda & 4\\ 4& 2-\lambda \end{array}\right) \end{gathered}$$

Find that the matrix$M-\lambda I$ is just equal toM with$\lambda$subtracted from each element on the main diagonal elements.

Computing $det\left(M-\lambda I\right)$implies that

$$\begin{gathered} det\left(M-\lambda I\right)=\left|M-\lambda I\right| \\ =\left|\begin{array}{cc}8-\lambda & 4\\ 4& 2-\lambda \end{array}\right| \\ =\left(8-\lambda \right)\left(2-\lambda \right)-\left(4\right)\left(4\right) \\ =16-8\lambda -2\lambda +{\lambda }^2-16 \\ ={\lambda }^2-10\lambda \end{gathered}$$

Thus, we have

$det(M-\lambda I)={\lambda }^2-10\lambda$

In the next step find the solution of $det(M-\lambda I)=0$(of the quadratic equation ${\lambda }^2-10\lambda =0$) as follows

The roots of the quadratic equation

${\lambda }^2-10\lambda =0$Implies that

$\lambda (\lambda -10)=0$

Find that

$\lambda =0$and

$\lambda =0$

Therefore, the eigenvalues of a square matrixM are $\lambda =0,10$.

In the next step, choose the principal axes of the conic section as our reference axes as follows Assume that the axes$(x^{\text{'}},y^{\text{'}})$are rotated by an angle$\theta$from the original axes $(x,y)$.

Thus, using the relation defined by

localid="1659076067236" $\left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)C^{-1}MC\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=K$……… (3)

Here Cis an orthogonal rotational matrix defined by

$C=\left(\begin{array}{cc}\cos \left(\theta \right)& -\sin \left(\theta \right)\\ \sin \left(\theta \right)& \cos \left(\theta \right)\end{array}\right)$

If the matrix C is the matrix which diagonalizes M, then equation (3) represent the equation of the conic section relative to its principal axes.

Therefore, chooseC to be the matrix which diagonalizes M, found that

$$\begin{gathered} C^{-1}MC=D \\ =\left(\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right) \end{gathered}$$

Here D represent the diagonal matrix whose diagonal elements are the eigenvalues of the $2\times 2$ square matrix M andC represent an orthogonal rotational matrix whose columns are the components of the unit eigenvectors.

From the values of the eigenvalues of a square matrixM which are $\lambda =0,10$, deduce that

$$\begin{gathered} C^{-1}MC=\left(\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right) \\ =\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right) \end{gathered}$$

Therefore, the conic section quadratic equation relative to the principal axes$\left(x^{\text{'}},y^{\text{'}}\right)$ becomes

$$\begin{gathered} \left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)C^{-1}MC\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=K \\ \Downarrow \\ \left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35 \end{gathered}$$

Thus, we finally have shown that the conic section quadratic equation relative to the principal axes $\left(x^{\text{'}},y^{\text{'}}\right)$in a matrix form is

$\left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35$……… (4)

Now, simplify the following equation (4) defined in a matrix form multiplied out gives the central conic section quadratic equation as follows

Suppose that the matrix $M_{12}$having 1 row and 2 columns, the matrix $M_{22}$having 2 rows and 2 columns, and the matrix $M_{21}$having 2 rows and 1 column given by

$M_{12}=\left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right),M_{22}=\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)$

and

$M_{21}=\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)$

Firstly, the product$M_{12}{M}_{22}$ between matrix$M_{12}$ having 1 row and 2 columns and matrix$M_{22}$ having 2 rows and 2 columns is the matrix having 1 row and 2 columns given by

$$\begin{gathered} M_{12}{M}_{22}={\left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)}_{1\times 2}{\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)}_{2\times 2} \\ =\left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right) \\ =\left(\left(x^{\text{'}}\right)\left(0\right)+\left(y^{\text{'}}\right)\left(0\right)\left(x^{\text{'}}\right)\left(0\right)+\left(y^{\text{'}}\right)\left(10\right)\right) \\ =\left(\begin{array}{cc}0& 10y^{\text{'}}\end{array}\right) \end{gathered}$$

Thus, shown that the following equation (4) defined in a matrix form becomes

$$\begin{gathered} \left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35 \\ \Downarrow \\ \left(\begin{array}{cc}0& 10y^{\text{'}}\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35 \end{gathered}$$

Suppose that the matrix$N_{12}$ having 1 row and 2 columns given by

$N_{12}=\left(\begin{array}{cc}0& 10y^{\text{'}}\end{array}\right)$

Finally, the product$N_{12}{M}_{21}$ between matrix$N_{12}$ having 1 row and 2 columns and matrix$M_{21}$ having 2 rows and 1 column is the matrix having 1 row and 1 column given by

$$\begin{gathered} N_{12}{M}_{21}={\left(\begin{array}{cc}0& 10y^{\text{'}}\end{array}\right)}_{1\times 2}{\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)}_{2\times 1} \\ =\left(\begin{array}{cc}0& 10y^{\text{'}}\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right) \\ =0x^{\text{'}2}+{10}^{\text{'}2}+0x^{\text{'}}{y}^{\text{'}} \\ =10y^{\text{'}2} \end{gathered}$$

Thus, finally the following equation (4) defined in a matrix form becomes

$$\begin{gathered} \left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35 \\ \left(\begin{array}{cc}0& 10y^{\text{'}}\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35 \\ \Downarrow \\ 10y^{\text{'}2}=35 \end{gathered}$$

Therefore, simplify the following equation (4) defined in a matrix form given by

$\left(\begin{array}{cc}{x}^{\text{'}}& y^{\text{'}}\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 10\end{array}\right)\left(\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\end{array}\right)=35$

Multiplied out gives the central conic section quadratic equation defined by $10y^{\text{'}2}=35$

Therefore, the conic section quadratic equation relative to the principal axes $(x^{\text{'}},y^{\text{'}})$is$10y^{\text{'}2}=35$