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:
- 25x^2-30xy+9y^2+68x+19=0
- 4x^2-16y^2-8x+96y-204=0
-
9x^2-4y^2-18x+16y-43=0
- 5x2+4xy+8y2−32x−56y+80=0
- Plot:
-
sin(x+y)=1.2*x
-
x^2+y^2-1
-
x^2+(y-cbrt(x^2))^2=1
- x^2+(y-(x^2)^(1/3))^2=1
-
Identical expressions
- two 5x^ two -3 zero xy+9y^2+68x+ nineteen =0
- 25x squared minus 30xy plus 9y squared plus 68x plus 19 equally 0
- two 5x to the power of two minus 3 zero xy plus 9y squared plus 68x plus nineteen equally 0
- 25x2-30xy+9y2+68x+19=0
- 25x²-30xy+9y²+68x+19=0
- 25x to the power of 2-30xy+9y to the power of 2+68x+19=0
- 25x^2-30xy+9y^2+68x+19=O
-
Similar expressions
- 25x^2-30xy+9y^2+68x-19=0
- 25x^2-30xy+9y^2-68x+19=0
- 25x^2+30xy+9y^2+68x+19=0
- 25x^2-30xy-9y^2+68x+19=0
25x^2-30xy+9y^2+68x+19=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 19 + 9*y + 25*x + 68*x - 30*x*y = 0
$$25 x^{2} - 30 x y + 68 x + 9 y^{2} + 19 = 0$$
25*x^2 - 30*x*y + 68*x + 9*y^2 + 19 = 0
Detail solution
Given line equation of 2-order:
$$25 x^{2} - 30 x y + 68 x + 9 y^{2} + 19 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -15$$
$$a_{13} = 34$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{33} = 19$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & -15\\-15 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{8}{15}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{8}{15} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{15}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{8}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{5 \sqrt{34}}{34}$$
$$\sin{\left(\phi \right)} = - \frac{3 \sqrt{34}}{34}$$
substitute coefficients
$$x' = \frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}$$
$$y' = - \frac{3 \sqrt{34} \tilde x}{34} + \frac{5 \sqrt{34} \tilde y}{34}$$
then the equation turns from
$$25 x'^{2} - 30 x' y' + 68 x' + 9 y'^{2} + 19 = 0$$
to
$$9 \left(- \frac{3 \sqrt{34} \tilde x}{34} + \frac{5 \sqrt{34} \tilde y}{34}\right)^{2} - 30 \left(- \frac{3 \sqrt{34} \tilde x}{34} + \frac{5 \sqrt{34} \tilde y}{34}\right) \left(\frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}\right) + 25 \left(\frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}\right)^{2} + 68 \left(\frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}\right) + 19 = 0$$
simplify
$$34 \tilde x^{2} + 10 \sqrt{34} \tilde x + 6 \sqrt{34} \tilde y + 19 = 0$$
$$\left(\sqrt{34} \tilde x + 5\right)^{2} = - 6 \sqrt{34} \tilde y + 6$$
$$\left(\tilde x + \frac{5 \sqrt{34}}{34}\right)^{2} = - \frac{3 \sqrt{34} \left(\tilde y - \frac{\sqrt{34}}{34}\right)}{17}$$
$$\tilde x'^{2} = - \frac{3 \sqrt{34} \tilde y'}{17}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{5 \sqrt{34}}{34} + 0 \left(- \frac{3 \sqrt{34}}{34}\right)$$
$$y_{0} = 0 \left(- \frac{3 \sqrt{34}}{34}\right) + 0 \frac{5 \sqrt{34}}{34}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{5 \sqrt{34}}{34}, \ - \frac{3 \sqrt{34}}{34}\right)$$
$$\vec e_2 = \left( \frac{3 \sqrt{34}}{34}, \ \frac{5 \sqrt{34}}{34}\right)$$
$$25 x^{2} - 30 x y + 68 x + 9 y^{2} + 19 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -15$$
$$a_{13} = 34$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{33} = 19$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & -15\\-15 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{8}{15}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{8}{15} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{15}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{8}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{5 \sqrt{34}}{34}$$
$$\sin{\left(\phi \right)} = - \frac{3 \sqrt{34}}{34}$$
substitute coefficients
$$x' = \frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}$$
$$y' = - \frac{3 \sqrt{34} \tilde x}{34} + \frac{5 \sqrt{34} \tilde y}{34}$$
then the equation turns from
$$25 x'^{2} - 30 x' y' + 68 x' + 9 y'^{2} + 19 = 0$$
to
$$9 \left(- \frac{3 \sqrt{34} \tilde x}{34} + \frac{5 \sqrt{34} \tilde y}{34}\right)^{2} - 30 \left(- \frac{3 \sqrt{34} \tilde x}{34} + \frac{5 \sqrt{34} \tilde y}{34}\right) \left(\frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}\right) + 25 \left(\frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}\right)^{2} + 68 \left(\frac{5 \sqrt{34} \tilde x}{34} + \frac{3 \sqrt{34} \tilde y}{34}\right) + 19 = 0$$
simplify
$$34 \tilde x^{2} + 10 \sqrt{34} \tilde x + 6 \sqrt{34} \tilde y + 19 = 0$$
$$\left(\sqrt{34} \tilde x + 5\right)^{2} = - 6 \sqrt{34} \tilde y + 6$$
$$\left(\tilde x + \frac{5 \sqrt{34}}{34}\right)^{2} = - \frac{3 \sqrt{34} \left(\tilde y - \frac{\sqrt{34}}{34}\right)}{17}$$
$$\tilde x'^{2} = - \frac{3 \sqrt{34} \tilde y'}{17}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{5 \sqrt{34}}{34} + 0 \left(- \frac{3 \sqrt{34}}{34}\right)$$
$$y_{0} = 0 \left(- \frac{3 \sqrt{34}}{34}\right) + 0 \frac{5 \sqrt{34}}{34}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{5 \sqrt{34}}{34}, \ - \frac{3 \sqrt{34}}{34}\right)$$
$$\vec e_2 = \left( \frac{3 \sqrt{34}}{34}, \ \frac{5 \sqrt{34}}{34}\right)$$
Invariants method
Given line equation of 2-order:
$$25 x^{2} - 30 x y + 68 x + 9 y^{2} + 19 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -15$$
$$a_{13} = 34$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{33} = 19$$
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
$$I_{1} = 34$$
$$I_{3} = \left|\begin{matrix}25 & -15 & 34\\-15 & 9 & 0\\34 & 0 & 19\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & -15\\-15 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 34$$
$$I_{2} = 0$$
$$I_{3} = -10404$$
$$I{\left(\lambda \right)} = \lambda^{2} - 34 \lambda$$
$$K_{2} = -510$$
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
$$6 \sqrt{34} \tilde x + 34 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{3 \sqrt{34}}{17} \tilde x$$
- reduced to canonical form
$$25 x^{2} - 30 x y + 68 x + 9 y^{2} + 19 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = -15$$
$$a_{13} = 34$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{33} = 19$$
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
$$I_{1} = 34$$
|25 -15|
I2 = | |
|-15 9 |$$I_{3} = \left|\begin{matrix}25 & -15 & 34\\-15 & 9 & 0\\34 & 0 & 19\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & -15\\-15 & 9 - \lambda\end{matrix}\right|$$
|25 34| |9 0 |
K2 = | | + | |
|34 19| |0 19|$$I_{1} = 34$$
$$I_{2} = 0$$
$$I_{3} = -10404$$
$$I{\left(\lambda \right)} = \lambda^{2} - 34 \lambda$$
$$K_{2} = -510$$
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
$$6 \sqrt{34} \tilde x + 34 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{3 \sqrt{34}}{17} \tilde x$$
- reduced to canonical form