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How to use it?
- Canonical form:
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9x^2-24xy+16y^2+18x+226y+209=0
- sin(x)/x
- 13
- 9x²-81y²=729
- Plot:
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x^2+(y+cbrt(x^2))^2=1
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(-x^2)*y^3+(x^2+y^2-1)^3=0
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sin(x^e)+cos(y^e)=1
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Identical expressions
- 7x^(two)+6xy-y^(two)+ twenty-eight x+12y+28= zero
- 7x to the power of (2) plus 6xy minus y to the power of (2) plus 28x plus 12y plus 28 equally 0
- 7x to the power of (two) plus 6xy minus y to the power of (two) plus twenty minus eight x plus 12y plus 28 equally zero
- 7x(2)+6xy-y(2)+28x+12y+28=0
- 7x2+6xy-y2+28x+12y+28=0
- 7x^2+6xy-y^2+28x+12y+28=0
- 7x^(2)+6xy-y^(2)+28x+12y+28=O
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Similar expressions
- 7x^(2)-6xy-y^(2)+28x+12y+28=0
- 7x^(2)+6xy-y^(2)+28x+12y-28=0
- 7x^(2)+6xy-y^(2)-28x+12y+28=0
- 7x^(2)+6xy+y^(2)+28x+12y+28=0
- 7x^(2)+6xy-y^(2)+28x-12y+28=0
7x^(2)+6xy-y^(2)+28x+12y+28=0 canonical form
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2 2 28 - y + 7*x + 12*y + 28*x + 6*x*y = 0
$$7 x^{2} + 6 x y - y^{2} + 28 x + 12 y + 28 = 0$$
7*x^2 + 6*x*y + 28*x - y^2 + 12*y + 28 = 0
Detail solution
Given line equation of 2-order:
$$7 x^{2} + 6 x y - y^{2} + 28 x + 12 y + 28 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 3$$
$$a_{13} = 14$$
$$a_{22} = -1$$
$$a_{23} = 6$$
$$a_{33} = 28$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}7 & 3\\3 & -1\end{matrix}\right|$$
$$\Delta = -16$$
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
$$7 x_{0} + 3 y_{0} + 14 = 0$$
$$3 x_{0} - y_{0} + 6 = 0$$
then
$$x_{0} = -2$$
$$y_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 14 x_{0} + 6 y_{0} + 28$$
$$a'_{33} = 0$$
then The equation is transformed to
$$7 x'^{2} + 6 x' y' - y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{4}{3}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{10}}{10}$$
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
$$7 x'^{2} + 6 x' y' - y'^{2} = 0$$
to
$$- \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} + 6 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) + 7 \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right)^{2} = 0$$
simplify
$$8 \tilde x^{2} - 2 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{3 \sqrt{10}}{10}, \ \frac{\sqrt{10}}{10}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
$$7 x^{2} + 6 x y - y^{2} + 28 x + 12 y + 28 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 3$$
$$a_{13} = 14$$
$$a_{22} = -1$$
$$a_{23} = 6$$
$$a_{33} = 28$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}7 & 3\\3 & -1\end{matrix}\right|$$
$$\Delta = -16$$
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
$$7 x_{0} + 3 y_{0} + 14 = 0$$
$$3 x_{0} - y_{0} + 6 = 0$$
then
$$x_{0} = -2$$
$$y_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 14 x_{0} + 6 y_{0} + 28$$
$$a'_{33} = 0$$
then The equation is transformed to
$$7 x'^{2} + 6 x' y' - y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{4}{3}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{10}}{10}$$
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
$$7 x'^{2} + 6 x' y' - y'^{2} = 0$$
to
$$- \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} + 6 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) + 7 \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right)^{2} = 0$$
simplify
$$8 \tilde x^{2} - 2 \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-2, 0)
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:
$$7 x^{2} + 6 x y - y^{2} + 28 x + 12 y + 28 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 3$$
$$a_{13} = 14$$
$$a_{22} = -1$$
$$a_{23} = 6$$
$$a_{33} = 28$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}7 & 3 & 14\\3 & -1 & 6\\14 & 6 & 28\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 7 & 3\\3 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = -16$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda - 16$$
$$K_{2} = -64$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 6 \lambda - 16 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = -2$$
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
$$8 \tilde x^{2} - 2 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0$$
- reduced to canonical form
$$7 x^{2} + 6 x y - y^{2} + 28 x + 12 y + 28 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 3$$
$$a_{13} = 14$$
$$a_{22} = -1$$
$$a_{23} = 6$$
$$a_{33} = 28$$
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
$$I_{1} = 6$$
|7 3 |
I2 = | |
|3 -1|$$I_{3} = \left|\begin{matrix}7 & 3 & 14\\3 & -1 & 6\\14 & 6 & 28\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 7 & 3\\3 & - \lambda - 1\end{matrix}\right|$$
|7 14| |-1 6 |
K2 = | | + | |
|14 28| |6 28|$$I_{1} = 6$$
$$I_{2} = -16$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda - 16$$
$$K_{2} = -64$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 6 \lambda - 16 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = -2$$
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
$$8 \tilde x^{2} - 2 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0$$
- reduced to canonical form
The graph