Mister Exam
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- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
-
How to use it?
- Canonical form:
- y*y-6*x-6*y-15=0
- x^2-6xy+10y^2=0
- x=2*y*y-12*y+14
- x^2-3xy-3y^2+2x+18y-62=0
- Plot:
-
x^2/2-2*x=y^2/2-y-3/2
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x=449.4*(y/100000-1)*(5/(1.4*(6*y/100000+1)))^(1/2)
- x-y=3
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x/4117+y^2-2=0
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Identical expressions
- x^ two -3xy-3y^ two +2x+18y- sixty-two = zero
- x squared minus 3xy minus 3y squared plus 2x plus 18y minus 62 equally 0
- x to the power of two minus 3xy minus 3y to the power of two plus 2x plus 18y minus sixty minus two equally zero
- x2-3xy-3y2+2x+18y-62=0
- x²-3xy-3y²+2x+18y-62=0
- x to the power of 2-3xy-3y to the power of 2+2x+18y-62=0
- x^2-3xy-3y^2+2x+18y-62=O
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Similar expressions
- x^2-3xy-3y^2+2x+18y+62=0
- x^2-3xy-3y^2+2x-18y-62=0
- x^2+3xy-3y^2+2x+18y-62=0
- x^2-3xy+3y^2+2x+18y-62=0
- x^2-3xy-3y^2-2x+18y-62=0
x^2-3xy-3y^2+2x+18y-62=0 canonical form
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The solution
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2 2 -62 + x - 3*y + 2*x + 18*y - 3*x*y = 0
$$x^{2} - 3 x y + 2 x - 3 y^{2} + 18 y - 62 = 0$$
x^2 - 3*x*y + 2*x - 3*y^2 + 18*y - 62 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 3 x y + 2 x - 3 y^{2} + 18 y - 62 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = - \frac{3}{2}$$
$$a_{13} = 1$$
$$a_{22} = -3$$
$$a_{23} = 9$$
$$a_{33} = -62$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & - \frac{3}{2}\\- \frac{3}{2} & -3\end{matrix}\right|$$
$$\Delta = - \frac{21}{4}$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$x_{0} - \frac{3 y_{0}}{2} + 1 = 0$$
$$- \frac{3 x_{0}}{2} - 3 y_{0} + 9 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = 2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = x_{0} + 9 y_{0} - 62$$
$$a'_{33} = -42$$
then equation turns into
$$x'^{2} - 3 x' y' - 3 y'^{2} - 42 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{4}{3}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{10}}{10}$$
substitute coefficients
$$x' = \frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}$$
$$y' = - \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}$$
then the equation turns from
$$x'^{2} - 3 x' y' - 3 y'^{2} - 42 = 0$$
to
$$- 3 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} - 3 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right) + \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right)^{2} - 42 = 0$$
simplify
$$\frac{3 \tilde x^{2}}{2} - \frac{7 \tilde y^{2}}{2} - 42 = 0$$
$$- \frac{3 \tilde x^{2}}{2} + \frac{7 \tilde y^{2}}{2} + 42 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{28} - \frac{\tilde y^{2}}{12} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ - \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
$$x^{2} - 3 x y + 2 x - 3 y^{2} + 18 y - 62 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = - \frac{3}{2}$$
$$a_{13} = 1$$
$$a_{22} = -3$$
$$a_{23} = 9$$
$$a_{33} = -62$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & - \frac{3}{2}\\- \frac{3}{2} & -3\end{matrix}\right|$$
$$\Delta = - \frac{21}{4}$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$x_{0} - \frac{3 y_{0}}{2} + 1 = 0$$
$$- \frac{3 x_{0}}{2} - 3 y_{0} + 9 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = 2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = x_{0} + 9 y_{0} - 62$$
$$a'_{33} = -42$$
then equation turns into
$$x'^{2} - 3 x' y' - 3 y'^{2} - 42 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{4}{3}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{10}}{10}$$
substitute coefficients
$$x' = \frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}$$
$$y' = - \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}$$
then the equation turns from
$$x'^{2} - 3 x' y' - 3 y'^{2} - 42 = 0$$
to
$$- 3 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} - 3 \left(- \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right) + \left(\frac{3 \sqrt{10} \tilde x}{10} + \frac{\sqrt{10} \tilde y}{10}\right)^{2} - 42 = 0$$
simplify
$$\frac{3 \tilde x^{2}}{2} - \frac{7 \tilde y^{2}}{2} - 42 = 0$$
$$- \frac{3 \tilde x^{2}}{2} + \frac{7 \tilde y^{2}}{2} + 42 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{28} - \frac{\tilde y^{2}}{12} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, 2)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ - \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 3 x y + 2 x - 3 y^{2} + 18 y - 62 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = - \frac{3}{2}$$
$$a_{13} = 1$$
$$a_{22} = -3$$
$$a_{23} = 9$$
$$a_{33} = -62$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = -2$$
$$I_{3} = \left|\begin{matrix}1 & - \frac{3}{2} & 1\\- \frac{3}{2} & -3 & 9\\1 & 9 & -62\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{3}{2}\\- \frac{3}{2} & - \lambda - 3\end{matrix}\right|$$
$$I_{1} = -2$$
$$I_{2} = - \frac{21}{4}$$
$$I_{3} = \frac{441}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda - \frac{21}{4}$$
$$K_{2} = 42$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 2 \lambda - \frac{21}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = - \frac{7}{2}$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\frac{3 \tilde x^{2}}{2} - \frac{7 \tilde y^{2}}{2} - 42 = 0$$
$$\frac{\tilde x^{2}}{28} - \frac{\tilde y^{2}}{12} = 1$$
- reduced to canonical form
$$x^{2} - 3 x y + 2 x - 3 y^{2} + 18 y - 62 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = - \frac{3}{2}$$
$$a_{13} = 1$$
$$a_{22} = -3$$
$$a_{23} = 9$$
$$a_{33} = -62$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|substitute coefficients
$$I_{1} = -2$$
| 1 -3/2|
I2 = | |
|-3/2 -3 |$$I_{3} = \left|\begin{matrix}1 & - \frac{3}{2} & 1\\- \frac{3}{2} & -3 & 9\\1 & 9 & -62\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{3}{2}\\- \frac{3}{2} & - \lambda - 3\end{matrix}\right|$$
|1 1 | |-3 9 |
K2 = | | + | |
|1 -62| |9 -62|$$I_{1} = -2$$
$$I_{2} = - \frac{21}{4}$$
$$I_{3} = \frac{441}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda - \frac{21}{4}$$
$$K_{2} = 42$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 2 \lambda - \frac{21}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = - \frac{7}{2}$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\frac{3 \tilde x^{2}}{2} - \frac{7 \tilde y^{2}}{2} - 42 = 0$$
$$\frac{\tilde x^{2}}{28} - \frac{\tilde y^{2}}{12} = 1$$
- reduced to canonical form