Mister Exam
Other calculators
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- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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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:
- (x^2+y^2-1)^3-x^2*y^3=0
- 3cos(2x)-x+0,25=0
-
2,55y-1,55x=0
- 2*y-x=6
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Identical expressions
- two 5y^2+42x+144y= zero
- 25y squared plus 42x plus 144y equally 0
- two 5y squared plus 42x plus 144y equally zero
- 25y2+42x+144y=0
- 25y²+42x+144y=0
- 25y to the power of 2+42x+144y=0
- 25y^2+42x+144y=O
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Similar expressions
- 25y^2-42x+144y=0
- 25y^2+42x-144y=0
25y^2+42x+144y=0 canonical form
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The solution
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2 25*y + 42*x + 144*y = 0
$$42 x + 25 y^{2} + 144 y = 0$$
42*x + 25*y^2 + 144*y = 0
Detail solution
Given line equation of 2-order:
$$42 x + 25 y^{2} + 144 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 21$$
$$a_{22} = 25$$
$$a_{23} = 72$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(5 \tilde y + \frac{72}{5}\right)^{2} = \frac{5184}{25} - 42 \tilde x$$
$$\left(\tilde y + \frac{72}{25}\right)^{2} = \frac{5184}{625} - \frac{42 \tilde x}{25}$$
$$\tilde y'^{2} = \frac{5184}{625} - \frac{42 \tilde x}{25}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$42 x + 25 y^{2} + 144 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 21$$
$$a_{22} = 25$$
$$a_{23} = 72$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(5 \tilde y + \frac{72}{5}\right)^{2} = \frac{5184}{25} - 42 \tilde x$$
$$\left(\tilde y + \frac{72}{25}\right)^{2} = \frac{5184}{625} - \frac{42 \tilde x}{25}$$
$$\tilde y'^{2} = \frac{5184}{625} - \frac{42 \tilde x}{25}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$42 x + 25 y^{2} + 144 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 21$$
$$a_{22} = 25$$
$$a_{23} = 72$$
$$a_{33} = 0$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 21\\0 & 25 & 72\\21 & 72 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 25 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = -11025$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = -5625$$
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
$$42 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{42 \tilde x}{25}$$
- reduced to canonical form
$$42 x + 25 y^{2} + 144 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 21$$
$$a_{22} = 25$$
$$a_{23} = 72$$
$$a_{33} = 0$$
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
$$I_{1} = 25$$
|0 0 |
I2 = | |
|0 25|$$I_{3} = \left|\begin{matrix}0 & 0 & 21\\0 & 25 & 72\\21 & 72 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 25 - \lambda\end{matrix}\right|$$
|0 21| |25 72|
K2 = | | + | |
|21 0 | |72 0 |$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = -11025$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = -5625$$
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
$$42 \tilde x + 25 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{42 \tilde x}{25}$$
- reduced to canonical form