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How to use it?
- Canonical form:
- x^2-y^2+z^2=0
- x^2-8xy+7y^2
- -17x^2+18xy+7y^2-38x-34y+11=0
- x^2-2x+3+2y=0
- Plot:
- z=3*x+6*y-x^2-x*y+y^2
- y=-x3+3*x-2
-
x^2-4xy+3y^2=6
- x^2+y^2=26*y
-
Identical expressions
- x^ two -8xy+7y^ two
- x squared minus 8xy plus 7y squared
- x to the power of two minus 8xy plus 7y to the power of two
- x2-8xy+7y2
- x²-8xy+7y²
- x to the power of 2-8xy+7y to the power of 2
-
Similar expressions
- x^2-8xy-7y^2
- x^2+8xy+7y^2
x^2-8xy+7y^2 canonical form
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2 2 x + 7*y - 8*x*y = 0
$$x^{2} - 8 x y + 7 y^{2} = 0$$
x^2 - 8*x*y + 7*y^2 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 8 x y + 7 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -4\\-4 & 7\end{matrix}\right|$$
$$\Delta = -9$$
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
$$x_{0} - 4 y_{0} = 0$$
$$- 4 x_{0} + 7 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - 8 x' y' + 7 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}$$
$$y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$x'^{2} - 8 x' y' + 7 y'^{2} = 0$$
to
$$7 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 8 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} = 0$$
simplify
$$- \tilde x^{2} + 9 \tilde y^{2} = 0$$
$$\tilde x^{2} - 9 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{3}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$x^{2} - 8 x y + 7 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -4\\-4 & 7\end{matrix}\right|$$
$$\Delta = -9$$
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
$$x_{0} - 4 y_{0} = 0$$
$$- 4 x_{0} + 7 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - 8 x' y' + 7 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}$$
$$y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$x'^{2} - 8 x' y' + 7 y'^{2} = 0$$
to
$$7 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 8 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} = 0$$
simplify
$$- \tilde x^{2} + 9 \tilde y^{2} = 0$$
$$\tilde x^{2} - 9 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{3}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 8 x y + 7 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}1 & -4 & 0\\-4 & 7 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -4\\-4 & 7 - \lambda\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda - 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
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
$$9 \tilde x^{2} - \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{3}\right)^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0$$
- reduced to canonical form
$$x^{2} - 8 x y + 7 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 8$$
|1 -4|
I2 = | |
|-4 7 |$$I_{3} = \left|\begin{matrix}1 & -4 & 0\\-4 & 7 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -4\\-4 & 7 - \lambda\end{matrix}\right|$$
|1 0| |7 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 8$$
$$I_{2} = -9$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda - 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
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
$$9 \tilde x^{2} - \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{3}\right)^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0$$
- reduced to canonical form