Mister Exam
xx+4xy+4yy-20x+10y-50 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 -50 + x - 20*x + 4*y + 10*y + 4*x*y = 0
$$x^{2} + 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
x^2 + 4*x*y - 20*x + 4*y^2 + 10*y - 50 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 2\\2 & 4\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$x'^{2} + 4 x' y' - 20 x' + 4 y'^{2} + 10 y' - 50 = 0$$
to
$$4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 10 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 50 = 0$$
simplify
$$- 10 \sqrt{5} \tilde x + 5 \tilde y^{2} - 50 = 0$$
$$10 \sqrt{5} \tilde x - 5 \tilde y^{2} + 50 = 0$$
$$5 \tilde y^{2} = 10 \sqrt{5} \tilde x + 50$$
$$\tilde y^{2} = 2 \sqrt{5} \left(\tilde x + \sqrt{5}\right)$$
$$\tilde y'^{2} = 2 \sqrt{5} \tilde x'$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \left(- \frac{\sqrt{5}}{5}\right)$$
$$y_{0} = 0 \left(- \frac{\sqrt{5}}{5}\right) + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$x^{2} + 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 2\\2 & 4\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$x'^{2} + 4 x' y' - 20 x' + 4 y'^{2} + 10 y' - 50 = 0$$
to
$$4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 10 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 50 = 0$$
simplify
$$- 10 \sqrt{5} \tilde x + 5 \tilde y^{2} - 50 = 0$$
$$10 \sqrt{5} \tilde x - 5 \tilde y^{2} + 50 = 0$$
$$5 \tilde y^{2} = 10 \sqrt{5} \tilde x + 50$$
$$\tilde y^{2} = 2 \sqrt{5} \left(\tilde x + \sqrt{5}\right)$$
$$\tilde y'^{2} = 2 \sqrt{5} \tilde x'$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \left(- \frac{\sqrt{5}}{5}\right)$$
$$y_{0} = 0 \left(- \frac{\sqrt{5}}{5}\right) + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$x^{2} + 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}1 & 2 & -10\\2 & 4 & 5\\-10 & 5 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2\\2 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = -625$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = -375$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$10 \sqrt{5} \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \sqrt{5} \tilde x$$
- reduced to canonical form
$$x^{2} + 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
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
$$I_{1} = 5$$
|1 2|
I2 = | |
|2 4|$$I_{3} = \left|\begin{matrix}1 & 2 & -10\\2 & 4 & 5\\-10 & 5 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2\\2 & 4 - \lambda\end{matrix}\right|$$
| 1 -10| |4 5 |
K2 = | | + | |
|-10 -50| |5 -50|$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = -625$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = -375$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$10 \sqrt{5} \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \sqrt{5} \tilde x$$
- reduced to canonical form