Mister Exam
Other calculators
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How to use it?
- Canonical form:
- x^2+4*y^2+8*z^2-16
- 3x^2+6x-15y+15=0
- 16x^2+24xy+9y^2+4x+3y=0
- 13x^2+18xy+37y^2
- Plot:
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x^2/4+y^2/9=1
-
x^2/36-y^2=1
- x^2-2=y
- x^2+y^2=9
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Identical expressions
- 16x^ two + two 4xy+9y^2+4x+3y= zero
- 16x squared plus 24xy plus 9y squared plus 4x plus 3y equally 0
- 16x to the power of two plus two 4xy plus 9y squared plus 4x plus 3y equally zero
- 16x2+24xy+9y2+4x+3y=0
- 16x²+24xy+9y²+4x+3y=0
- 16x to the power of 2+24xy+9y to the power of 2+4x+3y=0
- 16x^2+24xy+9y^2+4x+3y=O
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Similar expressions
- 16x^2+24xy+9y^2-4x+3y=0
- 16x^2+24xy+9y^2+4x-3y=0
- 16x^2-24xy+9y^2+4x+3y=0
- 16x^2+24xy-9y^2+4x+3y=0
16x^2+24xy+9y^2+4x+3y=0 canonical form
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The solution
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2 2 3*y + 4*x + 9*y + 16*x + 24*x*y = 0
$$16 x^{2} + 24 x y + 4 x + 9 y^{2} + 3 y = 0$$
16*x^2 + 24*x*y + 4*x + 9*y^2 + 3*y = 0
Detail solution
Given line equation of 2-order:
$$16 x^{2} + 24 x y + 4 x + 9 y^{2} + 3 y = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 12$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 12\\12 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7}{24}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}$$
$$y' = \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$16 x'^{2} + 24 x' y' + 4 x' + 9 y'^{2} + 3 y' = 0$$
to
$$9 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 3 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 16 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} + 4 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) = 0$$
simplify
$$25 \tilde x^{2} + 5 \tilde x = 0$$
$$5 \tilde x^{2} + \tilde x = 0$$
$$\left(\sqrt{5} \tilde x + \frac{\sqrt{5}}{10}\right)^{2} = \frac{1}{20}$$
$$\left(\tilde x + \frac{1}{10}\right)^{2} = \frac{1}{100}$$
$$\tilde x'^{2} = \frac{1}{100}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x + \frac{1}{10}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{\left(-1\right) 4}{5 \cdot 10} + \frac{0 \cdot 3}{5}$$
$$y_{0} = \frac{\left(-1\right) 3}{5 \cdot 10} + \frac{0 \cdot 4}{5}$$
$$x_{0} = - \frac{2}{25}$$
$$y_{0} = - \frac{3}{50}$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
$$16 x^{2} + 24 x y + 4 x + 9 y^{2} + 3 y = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 12$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 12\\12 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7}{24}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}$$
$$y' = \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$16 x'^{2} + 24 x' y' + 4 x' + 9 y'^{2} + 3 y' = 0$$
to
$$9 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 3 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 16 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} + 4 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) = 0$$
simplify
$$25 \tilde x^{2} + 5 \tilde x = 0$$
$$5 \tilde x^{2} + \tilde x = 0$$
$$\left(\sqrt{5} \tilde x + \frac{\sqrt{5}}{10}\right)^{2} = \frac{1}{20}$$
$$\left(\tilde x + \frac{1}{10}\right)^{2} = \frac{1}{100}$$
$$\tilde x'^{2} = \frac{1}{100}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x + \frac{1}{10}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{\left(-1\right) 4}{5 \cdot 10} + \frac{0 \cdot 3}{5}$$
$$y_{0} = \frac{\left(-1\right) 3}{5 \cdot 10} + \frac{0 \cdot 4}{5}$$
$$x_{0} = - \frac{2}{25}$$
$$y_{0} = - \frac{3}{50}$$
The center of canonical coordinate system at point O
(-2/25, -3/50)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$16 x^{2} + 24 x y + 4 x + 9 y^{2} + 3 y = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 12$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = 0$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}16 & 12 & 2\\12 & 9 & \frac{3}{2}\\2 & \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 12\\12 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = - \frac{25}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$25 \tilde y^{2} - \frac{1}{4} = 0$$
- reduced to canonical form
$$16 x^{2} + 24 x y + 4 x + 9 y^{2} + 3 y = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 12$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = 0$$
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
$$I_{1} = 25$$
|16 12|
I2 = | |
|12 9 |$$I_{3} = \left|\begin{matrix}16 & 12 & 2\\12 & 9 & \frac{3}{2}\\2 & \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 12\\12 & 9 - \lambda\end{matrix}\right|$$
|16 2| | 9 3/2|
K2 = | | + | |
|2 0| |3/2 0 |$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = - \frac{25}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$25 \tilde y^{2} - \frac{1}{4} = 0$$
None
- reduced to canonical form