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-
How to use it?
- Canonical form:
-
7x^2+16xy-23y^2-14x-16y=218
- 4x1x2+x2^2+4x2x3+2x3^2-4x1-2x2-5=0
- sqrt(16-x^2-y^2)=z
- y^2-16x-8y+32=0
- Plot:
-
sqrt(x)+sqrt(y)=1
-
tan(y)^2=tan(x)
-
-1-(36*(x^2-y^2+5*x+6)-(36*x+132)*(2*x+5))/((x^2-y^2+5*x+6)^2+4*x^2*y+25*y^2+20*x*y^2)=0
- tan(x-y+(1/10))-x*y=0
-
Identical expressions
- 7x^ two +16xy- two 3y^2-14x-16y= two hundred and eighteen
- 7x squared plus 16xy minus 23y squared minus 14x minus 16y equally 218
- 7x to the power of two plus 16xy minus two 3y squared minus 14x minus 16y equally two hundred and eighteen
- 7x2+16xy-23y2-14x-16y=218
- 7x²+16xy-23y²-14x-16y=218
- 7x to the power of 2+16xy-23y to the power of 2-14x-16y=218
-
Similar expressions
- 7x^2+16xy-23y^2+14x-16y=218
- 7x^2+16xy-23y^2-14x+16y=218
- 7x^2-16xy-23y^2-14x-16y=218
- 7x^2+16xy+23y^2-14x-16y=218
7x^2+16xy-23y^2-14x-16y=218 canonical form
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The solution
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2 2 -218 - 23*y - 16*y - 14*x + 7*x + 16*x*y = 0
$$7 x^{2} + 16 x y - 23 y^{2} - 14 x - 16 y - 218 = 0$$
7*x^2 + 16*x*y - 14*x - 23*y^2 - 16*y - 218 = 0
Detail solution
Given line equation of 2-order:
$$7 x^{2} + 16 x y - 23 y^{2} - 14 x - 16 y - 218 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 8$$
$$a_{13} = -7$$
$$a_{22} = -23$$
$$a_{23} = -8$$
$$a_{33} = -218$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}7 & 8\\8 & -23\end{matrix}\right|$$
$$\Delta = -225$$
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
$$7 x_{0} + 8 y_{0} - 7 = 0$$
$$8 x_{0} - 23 y_{0} - 8 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 7 x_{0} - 8 y_{0} - 218$$
$$a'_{33} = -225$$
then The equation is transformed to
$$7 x'^{2} + 16 x' y' - 23 y'^{2} - 225 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{15}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{15}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{8}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{15}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{4 \sqrt{17}}{17}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{17}}{17}$$
substitute coefficients
$$x' = \frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}$$
$$y' = \frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}$$
then the equation turns from
$$7 x'^{2} + 16 x' y' - 23 y'^{2} - 225 = 0$$
to
$$- 23 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right)^{2} + 16 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right) \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right) + 7 \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right)^{2} - 225 = 0$$
simplify
$$9 \tilde x^{2} - 25 \tilde y^{2} - 225 = 0$$
$$- 9 \tilde x^{2} + 25 \tilde y^{2} + 225 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{4 \sqrt{17}}{17}, \ \frac{\sqrt{17}}{17}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{17}}{17}, \ \frac{4 \sqrt{17}}{17}\right)$$
$$7 x^{2} + 16 x y - 23 y^{2} - 14 x - 16 y - 218 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 8$$
$$a_{13} = -7$$
$$a_{22} = -23$$
$$a_{23} = -8$$
$$a_{33} = -218$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}7 & 8\\8 & -23\end{matrix}\right|$$
$$\Delta = -225$$
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
$$7 x_{0} + 8 y_{0} - 7 = 0$$
$$8 x_{0} - 23 y_{0} - 8 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 7 x_{0} - 8 y_{0} - 218$$
$$a'_{33} = -225$$
then The equation is transformed to
$$7 x'^{2} + 16 x' y' - 23 y'^{2} - 225 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{15}{8}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{15}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{8}{17}$$
$$\cos{\left(2 \phi \right)} = \frac{15}{17}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{4 \sqrt{17}}{17}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{17}}{17}$$
substitute coefficients
$$x' = \frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}$$
$$y' = \frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}$$
then the equation turns from
$$7 x'^{2} + 16 x' y' - 23 y'^{2} - 225 = 0$$
to
$$- 23 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right)^{2} + 16 \left(\frac{\sqrt{17} \tilde x}{17} + \frac{4 \sqrt{17} \tilde y}{17}\right) \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right) + 7 \left(\frac{4 \sqrt{17} \tilde x}{17} - \frac{\sqrt{17} \tilde y}{17}\right)^{2} - 225 = 0$$
simplify
$$9 \tilde x^{2} - 25 \tilde y^{2} - 225 = 0$$
$$- 9 \tilde x^{2} + 25 \tilde y^{2} + 225 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, 0)
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{4 \sqrt{17}}{17}, \ \frac{\sqrt{17}}{17}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{17}}{17}, \ \frac{4 \sqrt{17}}{17}\right)$$
Invariants method
Given line equation of 2-order:
$$7 x^{2} + 16 x y - 23 y^{2} - 14 x - 16 y - 218 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 8$$
$$a_{13} = -7$$
$$a_{22} = -23$$
$$a_{23} = -8$$
$$a_{33} = -218$$
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
$$I_{1} = -16$$
$$I_{3} = \left|\begin{matrix}7 & 8 & -7\\8 & -23 & -8\\-7 & -8 & -218\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 7 & 8\\8 & - \lambda - 23\end{matrix}\right|$$
$$I_{1} = -16$$
$$I_{2} = -225$$
$$I_{3} = 50625$$
$$I{\left(\lambda \right)} = \lambda^{2} + 16 \lambda - 225$$
$$K_{2} = 3375$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} + 16 \lambda - 225 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = -25$$
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
$$9 \tilde x^{2} - 25 \tilde y^{2} - 225 = 0$$
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
$$7 x^{2} + 16 x y - 23 y^{2} - 14 x - 16 y - 218 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 8$$
$$a_{13} = -7$$
$$a_{22} = -23$$
$$a_{23} = -8$$
$$a_{33} = -218$$
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
$$I_{1} = -16$$
|7 8 |
I2 = | |
|8 -23|$$I_{3} = \left|\begin{matrix}7 & 8 & -7\\8 & -23 & -8\\-7 & -8 & -218\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 7 & 8\\8 & - \lambda - 23\end{matrix}\right|$$
|7 -7 | |-23 -8 |
K2 = | | + | |
|-7 -218| |-8 -218|$$I_{1} = -16$$
$$I_{2} = -225$$
$$I_{3} = 50625$$
$$I{\left(\lambda \right)} = \lambda^{2} + 16 \lambda - 225$$
$$K_{2} = 3375$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} + 16 \lambda - 225 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = -25$$
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
$$9 \tilde x^{2} - 25 \tilde y^{2} - 225 = 0$$
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The graph