Mister Exam

# 16x^2-9y^2+64x+54y+127=0 canonical form

x: [, ]
y: [, ]
z: [, ]

#### Quality:

(Number of points on the axis)

### The solution

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         2       2
127 - 9*y  + 16*x  + 54*y + 64*x = 0
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
16*x^2 + 64*x - 9*y^2 + 54*y + 127 = 0
Detail solution
Given line equation of 2-order:
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = 32$$
$$a_{22} = -9$$
$$a_{23} = 27$$
$$a_{33} = 127$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -144$$
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
$$16 x_{0} + 32 = 0$$
$$27 - 9 y_{0} = 0$$
then
$$x_{0} = -2$$
$$y_{0} = 3$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 32 x_{0} + 27 y_{0} + 127$$
$$a'_{33} = 144$$
then equation turns into
$$16 x'^{2} - 9 y'^{2} + 144 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{16} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-2, 3)

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:
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = 32$$
$$a_{22} = -9$$
$$a_{23} = 27$$
$$a_{33} = 127$$
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
$$I_{1} = 7$$
     |16  0 |
I2 = |      |
|0   -9|

$$I_{3} = \left|\begin{matrix}16 & 0 & 32\\0 & -9 & 27\\32 & 27 & 127\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 0\\0 & - \lambda - 9\end{matrix}\right|$$
     |16  32 |   |-9  27 |
K2 = |       | + |       |
|32  127|   |27  127|

$$I_{1} = 7$$
$$I_{2} = -144$$
$$I_{3} = -20736$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda - 144$$
$$K_{2} = -864$$
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:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 7 \lambda - 144 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = -9$$
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
$$16 \tilde x^{2} - 9 \tilde y^{2} + 144 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{16} = -1$$
- reduced to canonical form
To see a detailed solution - share to all your student friends
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share to all your student friends: