Answer to Question #203160 in Analytic Geometry for rama

Question

What kind of conic section the following quadratic form represents and transform it to principal axes.

4x21 + 6x1x2 − 4x22 = 10.

Answer 1

We have $Q=x^TAx,$ where

$$A=\begin{bmatrix} 4 & 3 \\ 3 & -4 \end{bmatrix}, x=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

$$\begin{vmatrix} 4-\lambda & 3 \\ 3 & -4-\lambda \end{vmatrix}=0$$

$$(4-\lambda)(-4-\lambda)-9=0$$

$$\lambda^2-25=0$$

$\lambda_1=-5, \lambda_2=5$

These are the eigenvalues.

Hence

$$Q=-5y_1^2+5y_2^2$$

We see that $Q=10$ represents the hyperbola $-5y_1^2+5y_2^2=10$

$\lambda=-5$

$$\begin{bmatrix} 4-\lambda & 3 \\ 3 & -4-\lambda \end{bmatrix}=\begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix}$$

$R_1=R_1/3$

$$\begin{bmatrix} 1 & 1/3 \\ 3 & 1 \end{bmatrix}$$

$R_2=R_2-3R_1$

$$\begin{bmatrix} 1 & 1/3 \\ 0 & 0 \end{bmatrix}$$

If we take $x_2=t,$ then $x_1=-t/3.$

Thus

$$x=\begin{bmatrix} -t/3\\ t \end{bmatrix}=\begin{bmatrix} -1/3\\ 1 \end{bmatrix}t$$

$\lambda=5$

$$\begin{bmatrix} 4-\lambda & 3 \\ 3 & -4-\lambda \end{bmatrix}=\begin{bmatrix} -1 & 3 \\ 3 & -9 \end{bmatrix}$$

$R_1=-R_1$

$$\begin{bmatrix} 1 & -3 \\ 3 & -9 \end{bmatrix}$$

$R_2=R_2-3R_1$

$$\begin{bmatrix} 1 & -3 \\ 0 & 0 \end{bmatrix}$$

If we take $x_2=t,$ then $x_1=3t.$

Thus

$$x=\begin{bmatrix} 3t\\ t \end{bmatrix}=\begin{bmatrix} 3\\ 1 \end{bmatrix}t$$

$$\dfrac{3}{\sqrt{10}}\begin{bmatrix} -1/3\\ 1 \end{bmatrix}=\begin{bmatrix} -1/\sqrt{10}\\ 3/\sqrt{10} \end{bmatrix}$$

$$\dfrac{1}{\sqrt{10}}\begin{bmatrix} 3\\ 1 \end{bmatrix}=\begin{bmatrix} 3/\sqrt{10}\\ 1/\sqrt{10} \end{bmatrix}$$

Hence

$$x=\begin{bmatrix} -1/\sqrt{10} & 3/\sqrt{10} \\ 3/\sqrt{10} & 1/\sqrt{10} \end{bmatrix}\begin{bmatrix} y_1\\ y_2 \end{bmatrix}$$

$$x_1=-\dfrac{1}{\sqrt{10}}y_1+\dfrac{3}{\sqrt{10}}y_2$$

$$x_2=\dfrac{3}{\sqrt{10}}y_1+\dfrac{1}{\sqrt{10}}y_2$$