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- 5x^2–6y^2–20x–36y–64=0
- 6*x-7*y-4*z-2
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Identical expressions
- x^ two / one hundred and sixty-nine +y^ two / one hundred and forty-four = one
- x squared divide by 169 plus y squared divide by 144 equally 1
- x to the power of two divide by one hundred and sixty minus nine plus y to the power of two divide by one hundred and forty minus four equally one
- x2/169+y2/144=1
- x²/169+y²/144=1
- x to the power of 2/169+y to the power of 2/144=1
- x^2 divide by 169+y^2 divide by 144=1
-
Similar expressions
- x^2/169-y^2/144=1
x^2/169+y^2/144=1 canonical form
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The solution
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2 2
y x
-1 + --- + --- = 0
144 169
$$\frac{x^{2}}{169} + \frac{y^{2}}{144} - 1 = 0$$
x^2/169 + y^2/144 - 1 = 0
Detail solution
Given line equation of 2-order:
$$\frac{x^{2}}{169} + \frac{y^{2}}{144} - 1 = 0$$
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
$$a_{11} = \frac{1}{169}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{144}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}\frac{1}{169} & 0\\0 & \frac{1}{144}\end{matrix}\right|$$
$$\Delta = \frac{1}{24336}$$
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
$$\frac{x_{0}}{169} = 0$$
$$\frac{y_{0}}{144} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = -1$$
$$a'_{33} = -1$$
then The equation is transformed to
$$\frac{x'^{2}}{169} + \frac{y'^{2}}{144} - 1 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{13^{2}} + \frac{\tilde y^{2}}{12^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$\frac{x^{2}}{169} + \frac{y^{2}}{144} - 1 = 0$$
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
$$a_{11} = \frac{1}{169}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{144}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}\frac{1}{169} & 0\\0 & \frac{1}{144}\end{matrix}\right|$$
$$\Delta = \frac{1}{24336}$$
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
$$\frac{x_{0}}{169} = 0$$
$$\frac{y_{0}}{144} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = -1$$
$$a'_{33} = -1$$
then The equation is transformed to
$$\frac{x'^{2}}{169} + \frac{y'^{2}}{144} - 1 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{13^{2}} + \frac{\tilde y^{2}}{12^{2}} = 1$$
- reduced to canonical form
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:
$$\frac{x^{2}}{169} + \frac{y^{2}}{144} - 1 = 0$$
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
$$a_{11} = \frac{1}{169}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{144}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = \frac{313}{24336}$$
$$I_{3} = \left|\begin{matrix}\frac{1}{169} & 0 & 0\\0 & \frac{1}{144} & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \frac{1}{169} & 0\\0 & - \lambda + \frac{1}{144}\end{matrix}\right|$$
$$I_{1} = \frac{313}{24336}$$
$$I_{2} = \frac{1}{24336}$$
$$I_{3} = - \frac{1}{24336}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{313 \lambda}{24336} + \frac{1}{24336}$$
$$K_{2} = - \frac{313}{24336}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - \frac{313 \lambda}{24336} + \frac{1}{24336} = 0$$
Solve this equation
$$\lambda_{1} = \frac{1}{144}$$
$$\lambda_{2} = \frac{1}{169}$$
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
$$\frac{\tilde x^{2}}{144} + \frac{\tilde y^{2}}{169} - 1 = 0$$
$$\frac{\tilde x^{2}}{12^{2}} + \frac{\tilde y^{2}}{13^{2}} = 1$$
- reduced to canonical form
$$\frac{x^{2}}{169} + \frac{y^{2}}{144} - 1 = 0$$
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
$$a_{11} = \frac{1}{169}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{144}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = \frac{313}{24336}$$
|1/169 0 |
I2 = | |
| 0 1/144|$$I_{3} = \left|\begin{matrix}\frac{1}{169} & 0 & 0\\0 & \frac{1}{144} & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \frac{1}{169} & 0\\0 & - \lambda + \frac{1}{144}\end{matrix}\right|$$
|1/169 0 | |1/144 0 |
K2 = | | + | |
| 0 -1| | 0 -1|$$I_{1} = \frac{313}{24336}$$
$$I_{2} = \frac{1}{24336}$$
$$I_{3} = - \frac{1}{24336}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{313 \lambda}{24336} + \frac{1}{24336}$$
$$K_{2} = - \frac{313}{24336}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - \frac{313 \lambda}{24336} + \frac{1}{24336} = 0$$
Solve this equation
$$\lambda_{1} = \frac{1}{144}$$
$$\lambda_{2} = \frac{1}{169}$$
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
$$\frac{\tilde x^{2}}{144} + \frac{\tilde y^{2}}{169} - 1 = 0$$
$$\frac{\tilde x^{2}}{12^{2}} + \frac{\tilde y^{2}}{13^{2}} = 1$$
- reduced to canonical form
The graph