Mister Exam
Other calculators
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- 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:
-
x^2+8y^2+8x+14y+5=0
- 14x^2-7y^2+20xy-8x-34y-35
- (y+3)^2=4x-2
- 5x^2+y^2+10x-6y-10z+14=0
- Plot:
-
x^2*y=x^2-y^2
-
-0.47y-1.02y^3+0.5ln((1+y)/(1-y))=x(1-exp(2.2*x/2))
- x^2+y^2=16
-
ln((y+x)/(x+3))=1+(1/(x+y))
-
Identical expressions
- 14x^ two -7y^ two +20xy-8x-34y- thirty-five
- 14x squared minus 7y squared plus 20xy minus 8x minus 34y minus 35
- 14x to the power of two minus 7y to the power of two plus 20xy minus 8x minus 34y minus thirty minus five
- 14x2-7y2+20xy-8x-34y-35
- 14x²-7y²+20xy-8x-34y-35
- 14x to the power of 2-7y to the power of 2+20xy-8x-34y-35
-
Similar expressions
- 14x^2-7y^2+20xy-8x+34y-35
- 14x^2+7y^2+20xy-8x-34y-35
- 14x^2-7y^2+20xy-8x-34y+35
- 14x^2-7y^2-20xy-8x-34y-35
- 14x^2-7y^2+20xy+8x-34y-35
14x^2-7y^2+20xy-8x-34y-35 canonical form
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The solution
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2 2 -35 - 34*y - 8*x - 7*y + 14*x + 20*x*y = 0
$$14 x^{2} + 20 x y - 8 x - 7 y^{2} - 34 y - 35 = 0$$
14*x^2 + 20*x*y - 8*x - 7*y^2 - 34*y - 35 = 0
Detail solution
Given line equation of 2-order:
$$14 x^{2} + 20 x y - 8 x - 7 y^{2} - 34 y - 35 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 10$$
$$a_{13} = -4$$
$$a_{22} = -7$$
$$a_{23} = -17$$
$$a_{33} = -35$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}14 & 10\\10 & -7\end{matrix}\right|$$
$$\Delta = -198$$
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
$$14 x_{0} + 10 y_{0} - 4 = 0$$
$$10 x_{0} - 7 y_{0} - 17 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -1$$
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
$$a'_{33} = - 4 x_{0} - 17 y_{0} - 35$$
$$a'_{33} = -22$$
then equation turns into
$$14 x'^{2} + 20 x' y' - 7 y'^{2} - 22 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{21}{20}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{21}{20} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{20}{29}$$
$$\cos{\left(2 \phi \right)} = \frac{21}{29}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{5 \sqrt{29}}{29}$$
$$\sin{\left(\phi \right)} = \frac{2 \sqrt{29}}{29}$$
substitute coefficients
$$x' = \frac{5 \sqrt{29} \tilde x}{29} - \frac{2 \sqrt{29} \tilde y}{29}$$
$$y' = \frac{2 \sqrt{29} \tilde x}{29} + \frac{5 \sqrt{29} \tilde y}{29}$$
then the equation turns from
$$14 x'^{2} + 20 x' y' - 7 y'^{2} - 22 = 0$$
to
$$- 7 \left(\frac{2 \sqrt{29} \tilde x}{29} + \frac{5 \sqrt{29} \tilde y}{29}\right)^{2} + 20 \left(\frac{2 \sqrt{29} \tilde x}{29} + \frac{5 \sqrt{29} \tilde y}{29}\right) \left(\frac{5 \sqrt{29} \tilde x}{29} - \frac{2 \sqrt{29} \tilde y}{29}\right) + 14 \left(\frac{5 \sqrt{29} \tilde x}{29} - \frac{2 \sqrt{29} \tilde y}{29}\right)^{2} - 22 = 0$$
simplify
$$18 \tilde x^{2} - 11 \tilde y^{2} - 22 = 0$$
$$- 18 \tilde x^{2} + 11 \tilde y^{2} + 22 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{11}{9}} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{5 \sqrt{29}}{29}, \ \frac{2 \sqrt{29}}{29}\right)$$
$$\vec e_2 = \left( - \frac{2 \sqrt{29}}{29}, \ \frac{5 \sqrt{29}}{29}\right)$$
$$14 x^{2} + 20 x y - 8 x - 7 y^{2} - 34 y - 35 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 10$$
$$a_{13} = -4$$
$$a_{22} = -7$$
$$a_{23} = -17$$
$$a_{33} = -35$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}14 & 10\\10 & -7\end{matrix}\right|$$
$$\Delta = -198$$
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
$$14 x_{0} + 10 y_{0} - 4 = 0$$
$$10 x_{0} - 7 y_{0} - 17 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -1$$
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
$$a'_{33} = - 4 x_{0} - 17 y_{0} - 35$$
$$a'_{33} = -22$$
then equation turns into
$$14 x'^{2} + 20 x' y' - 7 y'^{2} - 22 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{21}{20}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{21}{20} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{20}{29}$$
$$\cos{\left(2 \phi \right)} = \frac{21}{29}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{5 \sqrt{29}}{29}$$
$$\sin{\left(\phi \right)} = \frac{2 \sqrt{29}}{29}$$
substitute coefficients
$$x' = \frac{5 \sqrt{29} \tilde x}{29} - \frac{2 \sqrt{29} \tilde y}{29}$$
$$y' = \frac{2 \sqrt{29} \tilde x}{29} + \frac{5 \sqrt{29} \tilde y}{29}$$
then the equation turns from
$$14 x'^{2} + 20 x' y' - 7 y'^{2} - 22 = 0$$
to
$$- 7 \left(\frac{2 \sqrt{29} \tilde x}{29} + \frac{5 \sqrt{29} \tilde y}{29}\right)^{2} + 20 \left(\frac{2 \sqrt{29} \tilde x}{29} + \frac{5 \sqrt{29} \tilde y}{29}\right) \left(\frac{5 \sqrt{29} \tilde x}{29} - \frac{2 \sqrt{29} \tilde y}{29}\right) + 14 \left(\frac{5 \sqrt{29} \tilde x}{29} - \frac{2 \sqrt{29} \tilde y}{29}\right)^{2} - 22 = 0$$
simplify
$$18 \tilde x^{2} - 11 \tilde y^{2} - 22 = 0$$
$$- 18 \tilde x^{2} + 11 \tilde y^{2} + 22 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{11}{9}} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{5 \sqrt{29}}{29}, \ \frac{2 \sqrt{29}}{29}\right)$$
$$\vec e_2 = \left( - \frac{2 \sqrt{29}}{29}, \ \frac{5 \sqrt{29}}{29}\right)$$
Invariants method
Given line equation of 2-order:
$$14 x^{2} + 20 x y - 8 x - 7 y^{2} - 34 y - 35 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 10$$
$$a_{13} = -4$$
$$a_{22} = -7$$
$$a_{23} = -17$$
$$a_{33} = -35$$
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
$$I_{1} = 7$$
$$I_{3} = \left|\begin{matrix}14 & 10 & -4\\10 & -7 & -17\\-4 & -17 & -35\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}14 - \lambda & 10\\10 & - \lambda - 7\end{matrix}\right|$$
$$I_{1} = 7$$
$$I_{2} = -198$$
$$I_{3} = 4356$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda - 198$$
$$K_{2} = -550$$
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
$$\lambda^{2} - 7 \lambda - 198 = 0$$
$$\lambda_{1} = 18$$
$$\lambda_{2} = -11$$
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
$$18 \tilde x^{2} - 11 \tilde y^{2} - 22 = 0$$
$$\frac{\tilde x^{2}}{\frac{11}{9}} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form
$$14 x^{2} + 20 x y - 8 x - 7 y^{2} - 34 y - 35 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 10$$
$$a_{13} = -4$$
$$a_{22} = -7$$
$$a_{23} = -17$$
$$a_{33} = -35$$
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
$$I_{1} = 7$$
|14 10|
I2 = | |
|10 -7|$$I_{3} = \left|\begin{matrix}14 & 10 & -4\\10 & -7 & -17\\-4 & -17 & -35\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}14 - \lambda & 10\\10 & - \lambda - 7\end{matrix}\right|$$
|14 -4 | |-7 -17|
K2 = | | + | |
|-4 -35| |-17 -35|$$I_{1} = 7$$
$$I_{2} = -198$$
$$I_{3} = 4356$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda - 198$$
$$K_{2} = -550$$
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
$$\lambda^{2} - 7 \lambda - 198 = 0$$
$$\lambda_{1} = 18$$
$$\lambda_{2} = -11$$
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
$$18 \tilde x^{2} - 11 \tilde y^{2} - 22 = 0$$
$$\frac{\tilde x^{2}}{\frac{11}{9}} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form