Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
-
How to use it?
- Canonical form:
- y^2-z^2+2*x+(z-y)^2=0
- -z=2x^2+18y^2
- 5x^2+10y^2-50=0
- 25y^2+42x+144y=0
- Plot:
-
0.63=0.49+0.41*(1-exp(-0.17*x/(exp(-(y/(2805-y))^0.45)-x)))
- y^2-4*x+4*y-20=0
- (x-y^2)*((|z|)-1)=0
- (x^2+y^2)^2=a^2*x*y
-
Identical expressions
- xy+4y^ two
- xy plus 4y squared
- xy plus 4y to the power of two
- xy+4y2
- xy+4y²
- xy+4y to the power of 2
-
Similar expressions
- xy-4y^2
xy+4y^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x y + 4 y^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & \frac{1}{2}\\\frac{1}{2} & 4\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
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
$$\frac{y_{0}}{2} = 0$$
$$\frac{x_{0}}{2} + 4 y_{0} = 0$$
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
$$x' y' + 4 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = -4$$
then
$$\phi = - \frac{\operatorname{acot}{\left(4 \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{17}}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{4 \sqrt{17}}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} + \tilde y \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}$$
then the equation turns from
$$x' y' + 4 y'^{2} = 0$$
to
$$4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} + \tilde y \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}\right)^{2} + \left(- \tilde x \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} + \tilde y \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}\right) = 0$$
simplify
$$- \frac{8 \sqrt{17} \tilde x^{2}}{17} - \tilde x^{2} \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + 2 \tilde x^{2} - 8 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \frac{4 \sqrt{17} \tilde x \tilde y}{17} + \tilde y^{2} \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \frac{8 \sqrt{17} \tilde y^{2}}{17} + 2 \tilde y^{2} = 0$$
$$- \frac{\sqrt{17} \tilde x^{2}}{2} + 2 \tilde x^{2} + 2 \tilde y^{2} + \frac{\sqrt{17} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \frac{\sqrt{17}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \frac{\sqrt{17}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}\right)$$
$$\vec e_{2} = \left( \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}, \ \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}\right)$$
$$x y + 4 y^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & \frac{1}{2}\\\frac{1}{2} & 4\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
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
$$\frac{y_{0}}{2} = 0$$
$$\frac{x_{0}}{2} + 4 y_{0} = 0$$
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
$$x' y' + 4 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = -4$$
then
$$\phi = - \frac{\operatorname{acot}{\left(4 \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{17}}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{4 \sqrt{17}}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} + \tilde y \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}$$
then the equation turns from
$$x' y' + 4 y'^{2} = 0$$
to
$$4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} + \tilde y \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}\right)^{2} + \left(- \tilde x \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} + \tilde y \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}\right) = 0$$
simplify
$$- \frac{8 \sqrt{17} \tilde x^{2}}{17} - \tilde x^{2} \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + 2 \tilde x^{2} - 8 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \frac{4 \sqrt{17} \tilde x \tilde y}{17} + \tilde y^{2} \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}} \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}} + \frac{8 \sqrt{17} \tilde y^{2}}{17} + 2 \tilde y^{2} = 0$$
$$- \frac{\sqrt{17} \tilde x^{2}}{2} + 2 \tilde x^{2} + 2 \tilde y^{2} + \frac{\sqrt{17} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \frac{\sqrt{17}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \frac{\sqrt{17}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}\right)$$
$$\vec e_{2} = \left( \sqrt{\frac{1}{2} - \frac{2 \sqrt{17}}{17}}, \ \sqrt{\frac{2 \sqrt{17}}{17} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$x y + 4 y^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 4$$
$$I_{3} = \left|\begin{matrix}0 & \frac{1}{2} & 0\\\frac{1}{2} & 4 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & \frac{1}{2}\\\frac{1}{2} & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 4$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4 \lambda - \frac{1}{4}$$
$$K_{2} = 0$$
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
$$\lambda^{2} - 4 \lambda - \frac{1}{4} = 0$$
Solve this equation
$$\lambda_{1} = 2 - \frac{\sqrt{17}}{2}$$
$$\lambda_{2} = 2 + \frac{\sqrt{17}}{2}$$
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
$$\tilde x^{2} \cdot \left(2 - \frac{\sqrt{17}}{2}\right) + \tilde y^{2} \cdot \left(2 + \frac{\sqrt{17}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \frac{\sqrt{17}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \frac{\sqrt{17}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
$$x y + 4 y^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 4$$
| 0 1/2|
I2 = | |
|1/2 4 |$$I_{3} = \left|\begin{matrix}0 & \frac{1}{2} & 0\\\frac{1}{2} & 4 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & \frac{1}{2}\\\frac{1}{2} & 4 - \lambda\end{matrix}\right|$$
|0 0| |4 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 4$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4 \lambda - \frac{1}{4}$$
$$K_{2} = 0$$
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
$$\lambda^{2} - 4 \lambda - \frac{1}{4} = 0$$
Solve this equation
$$\lambda_{1} = 2 - \frac{\sqrt{17}}{2}$$
$$\lambda_{2} = 2 + \frac{\sqrt{17}}{2}$$
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
$$\tilde x^{2} \cdot \left(2 - \frac{\sqrt{17}}{2}\right) + \tilde y^{2} \cdot \left(2 + \frac{\sqrt{17}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \frac{\sqrt{17}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \frac{\sqrt{17}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The graph