Classifying conic section

Question

In my lecture notes, I have the following solution.

Find canonical form and classify the conic section $x^2+8y^2-4xy-8x+6y-5=0$.

$x^2+8y^2-4xy-8x+6y-5=\frac{4}{109}(x-2y-4)^2+\frac{16}{109}(y-\frac54)^2-1=0$. Letting $X:=x-2y-4, Y:=y-\frac54$ we get the canonical form and it is an equation of ellipse.

How can we choose $X:=x-2y-4, Y:=y-\frac54$ arbitrarily and decide the canonical form? I did not understand this way. What is the mathematical reason?

Thanks in advance.

Answer 1

Something is wrong in your version. The $ Q^T D Q = H $ version below says your polynomial is $$ (x-2y - 4)^2 + 4 (y - \frac{5}{4})^2 - \frac{109}{4} $$ I get it, you multiplied through by $\frac{4}{109}$

Meanwhile, there is no particular canonical form. One may complete some squares and get a diagonalized matrix as shown. Meanwhile, there is an algorithm that works, see the referenced question.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ \frac{ 13 }{ 2 } & \frac{ 5 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 8 & 3 \\ - 4 & 3 & - 5 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & \frac{ 13 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - \frac{ 109 }{ 4 } \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 4 & - \frac{ 5 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - \frac{ 109 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ 0 & 1 & - \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 8 & 3 \\ - 4 & 3 & - 5 \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 8 & 3 \\ - 4 & 3 & - 5 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 8 & 3 \\ - 4 & 3 & - 5 \\ \end{array} \right) $$

==============================================

$$ E_{1} = \left( \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & - 4 \\ 0 & 4 & - 5 \\ - 4 & - 5 & - 5 \\ \end{array} \right) $$

==============================================

$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & 2 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & - 5 \\ 0 & - 5 & - 21 \\ \end{array} \right) $$

==============================================

$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & 2 & \frac{ 13 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ 0 & 1 & - \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - \frac{ 109 }{ 4 } \\ \end{array} \right) $$

==============================================

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ \frac{ 13 }{ 2 } & \frac{ 5 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 8 & 3 \\ - 4 & 3 & - 5 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & \frac{ 13 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - \frac{ 109 }{ 4 } \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 4 & - \frac{ 5 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - \frac{ 109 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ 0 & 1 & - \frac{ 5 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 2 & - 4 \\ - 2 & 8 & 3 \\ - 4 & 3 & - 5 \\ \end{array} \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Answer 2

If you're trying to find the canonical form of $f(x,y)=0$ where $f$ is a quadratic polynomial, I suggest finding the Hessian matrix $H_f$ at the critical point of $f$. In fact, if $\bar{p}$ is a critical point of $f$, then $f(x,y)$ equals the following: $$[(x,y)-\bar{p}]H_f(\bar{p})[(x,y)-\bar{p}]^T+f(\bar{p})$$

Even further, if $(\tilde{x},\tilde{y})$ satisfies $$(x,y)-\bar{p}=\tilde{x}v_1 +\tilde{y}v_2$$ where $v_2,v_2$ are orthonormal eigenvectors of $H_f(\bar{p})$ with eigenvalues $\lambda_1,\lambda_2$ then we can think of $f$ as a function of the variables $\tilde{x},\tilde{y}$. $$f(\tilde{x},\tilde{y})=\lambda_1 \tilde{x}^2+\lambda_2 \tilde{y}^2 + f(\bar{p})$$

You should also interpret $\text{span}(v_1)+\bar{p}$ as your $\tilde{x}-$ axis and $\text{span}(v_2)+\bar{p}$ as your $\tilde{y}-$ axis. Your level set $f^{-1}(0)$ will be centered around these axes.

Answer 3

Explicitly, for simplicity first find the center $$(x-\frac{13}2)^2 - 4 (x-\frac{13}2) (y-\frac54) + 8 (y-\frac54)^2 - \frac{109}4=0$$ and then find the (positive; it's an ellipse!) eigenvalues for the corresponding matrix $\lambda_{1,2}=\frac{9 \pm \sqrt{65}}2.$ Sort them $0\leq\lambda_1<\lambda_2.$ The rotation angle $\tan{2\theta}=\frac47,$ or $\cos{\theta}=\sqrt{\frac12 + \frac{7}{2 \sqrt{65}}},$ $\sin{\theta}=2 \sqrt{\frac2{65 + 7 \sqrt{65}}}.$ That is $\theta\approx 0.26$ or $0.26r$ arc for any $r\geq 0$ or $14.9^{\circ}.$ I.e. the equation is

$$\lambda_1((x-\frac{13}2)\cos{\theta}+(y-\frac54)\sin{\theta}))^2+\lambda_2(-(x-\frac{13}2)\sin{\theta}+(y-\frac54)\cos{\theta})^2-\frac{109}4=0,$$ resulting in the canonical form $$\frac{((x-\frac{13}2)\cos{\theta}+(y-\frac54)y\sin{\theta}))^2}{a^2}+\frac{(-(x-\frac{13}2)\sin{\theta}+(y-\frac54)\cos{\theta})^2}{b^2}-1=0,$$ where the half-axes are $a=\sqrt{\frac{109}{4\lambda_1}}\approx 7.62,b=\sqrt{\frac{109}{4\lambda_2}}\approx 1.79.$

Answer 4

The canonical form of the conic

$x^2 + 8 y^2 - 4 xy -8 x + 6y - 5 = 0 \tag{1} $

can be obtained as follows. Let $r = [x, y]^T $, and

$ Q = \begin{bmatrix} 1 && -2 \\ -2 && 8 \end{bmatrix} , b = \begin{bmatrix} -8 \\ 6 \end{bmatrix} , c = -5 $

Then the matrix-vector version of $(1)$ is

$ r^T Q r + b^T r + c = 0 \tag{2} $

Clearly, $Q$ is invertible, so first calculate the center of the conic. It is given by

$ r_0 = - \dfrac{1}{2} Q^{-1} b = - \dfrac{1}{2} \cdot \dfrac{1}{4} \begin{bmatrix} 8 && 2 \\ 2 && 1 \end{bmatrix} \begin{bmatrix} - 8 \\ 6 \end{bmatrix} = \begin{bmatrix} 6.5 \\ 1.25 \end{bmatrix} \tag{3}$

With this $r_0$ , the equation becomes

$ (r - r_0)^T Q (r - r_0) = r_0^T Q r_0 - c = 27.25 =:d \tag{4}$

Secondly, diagonalize $Q$ as follows. Write

$ Q = R D R^T \tag{5}$

where $ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} \tag{6} $

where $ \theta = \dfrac{1}{2} \tan^{-1} \left( \dfrac{ 2 Q_{12} }{ Q_{11} - Q_{22} } \right) \tag{7} $

And

$D_{11} = Q_{11} \cos^2 \theta + Q_{22} \sin^2 \theta + 2 Q_{12} \sin \theta \cos \theta \tag{8a}$

$D_{11} = Q_{11} \sin^2 \theta + Q_{22} \cos^2 \theta - 2 Q_{12} \sin \theta \cos \theta \tag{8b}$

You'll find that

$R = \begin{bmatrix} 0.256668 && 0.9665 \\ -0.9665 && 0.256668 \end{bmatrix} \tag{9}$

$D = \begin{bmatrix} 8.531129 && 0 \\ 0 && 0.468871 \end{bmatrix} \tag{10}$

With this, the equation becomes

$ (r - r_0)^T R D R^T (r - r_0) = 17.25 \tag{11} $

Now, thirdly divide through by the right hand side to get

$ (r - r_0)^T R G R^T (r - r_0) = 1 \tag{12}$

where

$ G = \dfrac{D}{27.25} = \begin{bmatrix} 0.31306895 && 0 \\ 0 && 0.01720628 \end{bmatrix} \tag{13}$

The diagonal matrix $G$ gives the lengths of the semi-major and semi-minor axes

$\text{Semi-major} = \dfrac{1}{\sqrt{ 0.01720628} } = 7.623536939 \tag{14a}$

$ \text{Semi-minor} = \dfrac{1}{\sqrt{ 0.31306895}} = 1.787228174 \tag{14b} $

Define a new coordinate reference frame whose origin the point $r_0$ , and whose $x$ axis points along the first column of $R$ and whose $y$ axis points along the second column of $R$ , then if we call the new coordiante $p$, then it follows that two coordinates $r$ and $p$ are related by

$ r = r_0 + R p \tag{15}$

Substitute this into equation $(12)$, you get

$ p^T G p = 1 \tag{16}$

Since $G$ is diagonal, this is the canonical equation of the given ellipse in the coordinate vector $p$.