Answer on Question #73767 – Math – Analytic Geometry
QUESTION
Check whether or not the conicoid represented by
5 x 2 + 4 y 2 − 4 y z + 2 x z + 2 x − 4 y − 8 z + 2 = 0 5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 5 x 2 + 4 y 2 − 4 yz + 2 x z + 2 x − 4 y − 8 z + 2 = 0
is central or not. If it is, transform the equation by shifting the origin to the center. Else, change any one coefficient to make the equation that of a central conicoid.
SOLUTION
THEOREM 1
The origin p . O ( 0 , 0 , 0 ) p. O(0,0,0) p . O ( 0 , 0 , 0 )
a x 2 + b y 2 + c z 2 + 2 f y z + 2 g z x + 2 h x y + 2 u x + 2 v y + 2 w z + d = 0 ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 a x 2 + b y 2 + c z 2 + 2 f yz + 2 g z x + 2 h x y + 2 ux + 2 v y + 2 w z + d = 0
If and only if
u = v = w = 0 u = v = w = 0 u = v = w = 0
THEOREM 2
A conicoid S, given by equation
a x 2 + b y 2 + c z 2 + 2 f y z + 2 g z x + 2 h x y + 2 u x + 2 v y + 2 w z + d = 0 ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 a x 2 + b y 2 + c z 2 + 2 f yz + 2 g z x + 2 h x y + 2 ux + 2 v y + 2 w z + d = 0
has the point P ( x 0 , y 0 , z 0 ) P(x_0, y_0, z_0) P ( x 0 , y 0 , z 0 )
{ a x 0 + h y 0 + g z 0 + u = 0 h x 0 + b y 0 + f z 0 + v = 0 g x 0 + f y 0 + c z 0 + w = 0 \left\{ \begin{array}{l}
ax_0 + hy_0 + gz_0 + u = 0 \\
hx_0 + by_0 + fz_0 + v = 0 \\
gx_0 + fy_0 + cz_0 + w = 0
\end{array} \right. ⎩ ⎨ ⎧ a x 0 + h y 0 + g z 0 + u = 0 h x 0 + b y 0 + f z 0 + v = 0 g x 0 + f y 0 + c z 0 + w = 0
Using these two theorems, we can solve the problem posed.
We reduce the equation from the problem to the standard form:
5 x 2 + 4 y 2 − 4 y z + 2 x z + 2 x − 4 y − 8 z + 2 = 0 → 5 a x 2 + 4 b y 2 + 0 c z 2 + 2 ⋅ ( − 2 ) ⏟ f y z + 2 ⋅ ( 1 ) ⏟ g x z + 0 h x y + \begin{array}{l}
5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 \rightarrow \\
\frac{5}{a}x^2 + \frac{4}{b}y^2 + \frac{0}{c}z^2 + 2 \cdot \underbrace{(-2)}_{f}yz + 2 \cdot \underbrace{(1)}_{g}xz + \frac{0}{h}xy + \\
\end{array} 5 x 2 + 4 y 2 − 4 yz + 2 x z + 2 x − 4 y − 8 z + 2 = 0 → a 5 x 2 + b 4 y 2 + c 0 z 2 + 2 ⋅ f ( − 2 ) yz + 2 ⋅ g ( 1 ) x z + h 0 x y + + 2 ⋅ ( 1 ) ⏟ u x + 2 ⋅ ( − 2 ) ⏟ v y + 2 ⋅ ( − 4 ) ⏟ w z + 2 = 0 + 2 \cdot \underbrace {(1)} _ {u} x + 2 \cdot \underbrace {(- 2)} _ {v} y + 2 \cdot \underbrace {(- 4)} _ {w} z + 2 = 0 + 2 ⋅ u ( 1 ) x + 2 ⋅ v ( − 2 ) y + 2 ⋅ w ( − 4 ) z + 2 = 0
As we can see,
{ u = 1 ≠ 0 v = − 2 ≠ 0 → O ( 0 , 0 , 0 ) is not the center of this surface (by THEOREM 1) w = − 4 ≠ 0 \left\{ \begin{array}{l} u = 1 \neq 0 \\ v = - 2 \neq 0 \rightarrow O (0, 0, 0) \text{ is not the center of this surface (by THEOREM 1)} \\ w = - 4 \neq 0 \end{array} \right. ⎩ ⎨ ⎧ u = 1 = 0 v = − 2 = 0 → O ( 0 , 0 , 0 ) is not the center of this surface (by THEOREM 1) w = − 4 = 0
In order to find the center, it is necessary to solve the system
{ a x 0 + h y 0 + g z 0 + u = 0 h x 0 + b y 0 + f z 0 + v = 0 → ( THEOREM 2 ) g x 0 + f y 0 + c z 0 + w = 0 \left\{ \begin{array}{l} a x _ {0} + h y _ {0} + g z _ {0} + u = 0 \\ h x _ {0} + b y _ {0} + f z _ {0} + v = 0 \rightarrow (\text{THEOREM 2}) \\ g x _ {0} + f y _ {0} + c z _ {0} + w = 0 \end{array} \right. ⎩ ⎨ ⎧ a x 0 + h y 0 + g z 0 + u = 0 h x 0 + b y 0 + f z 0 + v = 0 → ( THEOREM 2 ) g x 0 + f y 0 + c z 0 + w = 0
In our case,
{ 5 x 0 + 0 y 0 + 1 z 0 + 1 = 0 0 x 0 + 4 y 0 + ( − 2 ) z 0 + ( − 2 ) = 0 1 x 0 + ( − 2 ) y 0 + 0 z 0 + ( − 4 ) = 0 → { 5 x 0 + 0 y 0 + 1 z 0 = − 1 0 x 0 + 4 y 0 + ( − 2 ) z 0 = 2 1 x 0 + ( − 2 ) y 0 + 0 z 0 = 4 \left\{ \begin{array}{l} 5 x _ {0} + 0 y _ {0} + 1 z _ {0} + 1 = 0 \\ 0 x _ {0} + 4 y _ {0} + (- 2) z _ {0} + (- 2) = 0 \\ 1 x _ {0} + (- 2) y _ {0} + 0 z _ {0} + (- 4) = 0 \end{array} \right. \rightarrow \left\{ \begin{array}{l} 5 x _ {0} + 0 y _ {0} + 1 z _ {0} = - 1 \\ 0 x _ {0} + 4 y _ {0} + (- 2) z _ {0} = 2 \\ 1 x _ {0} + (- 2) y _ {0} + 0 z _ {0} = 4 \end{array} \right. ⎩ ⎨ ⎧ 5 x 0 + 0 y 0 + 1 z 0 + 1 = 0 0 x 0 + 4 y 0 + ( − 2 ) z 0 + ( − 2 ) = 0 1 x 0 + ( − 2 ) y 0 + 0 z 0 + ( − 4 ) = 0 → ⎩ ⎨ ⎧ 5 x 0 + 0 y 0 + 1 z 0 = − 1 0 x 0 + 4 y 0 + ( − 2 ) z 0 = 2 1 x 0 + ( − 2 ) y 0 + 0 z 0 = 4
This system will be solved using Cramer's rule
(More information: https://en.wikipedia.org/wiki/Cramer%27s_rule)
Δ = ∣ 5 0 1 0 4 − 2 1 − 2 0 ∣ = = 5 ⋅ 4 ⋅ 0 + 0 ⋅ ( − 2 ) ⋅ 1 + 0 ⋅ ( − 2 ) ⋅ 1 − 1 ⋅ 1 ⋅ 4 − 5 ⋅ ( − 2 ) ⋅ ( − 2 ) − 0 ⋅ 0 ⋅ 0 = = 0 + 0 + 0 − 4 − 20 − 0 = − 24 \begin{array}{l} \Delta = \left| \begin{array}{c c c} 5 & 0 & 1 \\ 0 & 4 & - 2 \\ 1 & - 2 & 0 \end{array} \right| = \\ = 5 \cdot 4 \cdot 0 + 0 \cdot (- 2) \cdot 1 + 0 \cdot (- 2) \cdot 1 - 1 \cdot 1 \cdot 4 - 5 \cdot (- 2) \cdot (- 2) - 0 \cdot 0 \cdot 0 = \\ = 0 + 0 + 0 - 4 - 20 - 0 = - 24 \\ \end{array} Δ = ∣ ∣ 5 0 1 0 4 − 2 1 − 2 0 ∣ ∣ = = 5 ⋅ 4 ⋅ 0 + 0 ⋅ ( − 2 ) ⋅ 1 + 0 ⋅ ( − 2 ) ⋅ 1 − 1 ⋅ 1 ⋅ 4 − 5 ⋅ ( − 2 ) ⋅ ( − 2 ) − 0 ⋅ 0 ⋅ 0 = = 0 + 0 + 0 − 4 − 20 − 0 = − 24 Δ = − 24 \Delta = - 24 Δ = − 24 Δ x 0 = ∣ − 1 0 1 2 4 − 2 4 − 2 0 ∣ = \Delta_{x_0} = \left| \begin{array}{ccc} -1 & 0 & 1 \\ 2 & 4 & -2 \\ 4 & -2 & 0 \end{array} \right| = Δ x 0 = ∣ ∣ − 1 2 4 0 4 − 2 1 − 2 0 ∣ ∣ = = ( − 1 ) ⋅ 4 ⋅ 0 + 2 ⋅ ( − 2 ) ⋅ 1 + 0 ⋅ ( − 2 ) ⋅ 4 − 1 ⋅ 4 ⋅ 4 − ( − 1 ) ⋅ ( − 2 ) ⋅ ( − 2 ) − 2 ⋅ 0 ⋅ 0 = = (-1) \cdot 4 \cdot 0 + 2 \cdot (-2) \cdot 1 + 0 \cdot (-2) \cdot 4 - 1 \cdot 4 \cdot 4 - (-1) \cdot (-2) \cdot (-2) - 2 \cdot 0 \cdot 0 = = ( − 1 ) ⋅ 4 ⋅ 0 + 2 ⋅ ( − 2 ) ⋅ 1 + 0 ⋅ ( − 2 ) ⋅ 4 − 1 ⋅ 4 ⋅ 4 − ( − 1 ) ⋅ ( − 2 ) ⋅ ( − 2 ) − 2 ⋅ 0 ⋅ 0 = = 0 − 4 + 0 − 16 + 4 − 0 = − 16 = 0 - 4 + 0 - 16 + 4 - 0 = -16 = 0 − 4 + 0 − 16 + 4 − 0 = − 16 Δ x 0 = − 16 \boxed{\Delta_{x_0} = -16} Δ x 0 = − 16 x 0 = Δ x 0 Δ = − 16 − 24 = 2 ⋅ 8 3 ⋅ 8 = 2 3 → x 0 = 2 3 = 0. ( 6 ) ≈ 0.67 x_0 = \frac{\Delta_{x_0}}{\Delta} = \frac{-16}{-24} = \frac{2 \cdot 8}{3 \cdot 8} = \frac{2}{3} \rightarrow \boxed{x_0 = \frac{2}{3} = 0. (6) \approx 0.67} x 0 = Δ Δ x 0 = − 24 − 16 = 3 ⋅ 8 2 ⋅ 8 = 3 2 → x 0 = 3 2 = 0. ( 6 ) ≈ 0.67 Δ y 0 = ∣ 5 − 1 1 0 2 − 2 1 4 0 ∣ = \Delta_{y_0} = \left| \begin{array}{ccc} 5 & -1 & 1 \\ 0 & 2 & -2 \\ 1 & 4 & 0 \end{array} \right| = Δ y 0 = ∣ ∣ 5 0 1 − 1 2 4 1 − 2 0 ∣ ∣ = = 5 ⋅ 2 ⋅ 0 + ( − 1 ) ⋅ ( − 2 ) ⋅ 1 + 0 ⋅ 1 ⋅ 4 − 1 ⋅ 2 ⋅ 1 − 5 ⋅ 4 ⋅ ( − 2 ) − ( − 1 ) ⋅ 0 ⋅ 0 = = 5 \cdot 2 \cdot 0 + (-1) \cdot (-2) \cdot 1 + 0 \cdot 1 \cdot 4 - 1 \cdot 2 \cdot 1 - 5 \cdot 4 \cdot (-2) - (-1) \cdot 0 \cdot 0 = = 5 ⋅ 2 ⋅ 0 + ( − 1 ) ⋅ ( − 2 ) ⋅ 1 + 0 ⋅ 1 ⋅ 4 − 1 ⋅ 2 ⋅ 1 − 5 ⋅ 4 ⋅ ( − 2 ) − ( − 1 ) ⋅ 0 ⋅ 0 = = 0 + 2 + 0 − 2 + 40 − 0 = 40 = 0 + 2 + 0 - 2 + 40 - 0 = 40 = 0 + 2 + 0 − 2 + 40 − 0 = 40 Δ y 0 = 40 \boxed{\Delta_{y_0} = 40} Δ y 0 = 40 y 0 = Δ y 0 Δ = 40 − 24 = − 5 ⋅ 8 3 ⋅ 8 = − 5 3 → y 0 = − 5 3 = − 1. ( 6 ) ≈ − 1.67 y_0 = \frac{\Delta_{y_0}}{\Delta} = \frac{40}{-24} = -\frac{5 \cdot 8}{3 \cdot 8} = -\frac{5}{3} \rightarrow \boxed{y_0 = -\frac{5}{3} = -1. (6) \approx -1.67} y 0 = Δ Δ y 0 = − 24 40 = − 3 ⋅ 8 5 ⋅ 8 = − 3 5 → y 0 = − 3 5 = − 1. ( 6 ) ≈ − 1.67 Δ z 0 = ∣ 5 0 − 1 0 4 2 1 − 2 4 ∣ = \Delta_{z_0} = \left| \begin{array}{ccc} 5 & 0 & -1 \\ 0 & 4 & 2 \\ 1 & -2 & 4 \end{array} \right| = Δ z 0 = ∣ ∣ 5 0 1 0 4 − 2 − 1 2 4 ∣ ∣ = = 5 ⋅ 4 ⋅ 4 + ( − 1 ) ⋅ ( − 2 ) ⋅ 0 + 0 ⋅ 1 ⋅ 2 − 1 ⋅ 4 ⋅ ( − 1 ) − 5 ⋅ 2 ⋅ ( − 2 ) − 4 ⋅ 0 ⋅ 0 = = 5 \cdot 4 \cdot 4 + (-1) \cdot (-2) \cdot 0 + 0 \cdot 1 \cdot 2 - 1 \cdot 4 \cdot (-1) - 5 \cdot 2 \cdot (-2) - 4 \cdot 0 \cdot 0 = = 5 ⋅ 4 ⋅ 4 + ( − 1 ) ⋅ ( − 2 ) ⋅ 0 + 0 ⋅ 1 ⋅ 2 − 1 ⋅ 4 ⋅ ( − 1 ) − 5 ⋅ 2 ⋅ ( − 2 ) − 4 ⋅ 0 ⋅ 0 = = 80 − 0 + 0 + 4 + 20 − 0 = 104 = 80 - 0 + 0 + 4 + 20 - 0 = 104 = 80 − 0 + 0 + 4 + 20 − 0 = 104 Δ z 0 = 104 \boxed{\Delta_{z_0} = 104} Δ z 0 = 104 z 0 = Δ z 0 Δ = 104 − 24 = − 8 ⋅ 13 8 ⋅ 3 = − 13 3 → z 0 = − 13 3 = − 4. ( 3 ) ≈ − 4.33 z_0 = \frac{\Delta_{z_0}}{\Delta} = \frac{104}{-24} = -\frac{8 \cdot 13}{8 \cdot 3} = -\frac{13}{3} \rightarrow \boxed{z_0 = -\frac{13}{3} = -4. (3) \approx -4.33} z 0 = Δ Δ z 0 = − 24 104 = − 8 ⋅ 3 8 ⋅ 13 = − 3 13 → z 0 = − 3 13 = − 4. ( 3 ) ≈ − 4.33
Conclusion,
P ( 2 3 , − 5 3 , − 13 3 ) is the center of the conicoid P \left(\frac {2}{3}, - \frac {5}{3}, - \frac {13}{3}\right) \text{ is the center of the conicoid} P ( 3 2 , − 3 5 , − 3 13 ) is the center of the conicoid
In order for the conicoid to become central, it is necessary to apply a shift of the form
{ X = x − 2 3 Y = y + 5 3 Z = z + 13 3 → { x = X + 2 3 y = Y − 5 3 z = Z − 13 3 \left\{ \begin{array}{l} X = x - \frac {2}{3} \\ Y = y + \frac {5}{3} \\ Z = z + \frac {13}{3} \end{array} \right. \to \left\{ \begin{array}{l} x = X + \frac {2}{3} \\ y = Y - \frac {5}{3} \\ z = Z - \frac {13}{3} \end{array} \right. ⎩ ⎨ ⎧ X = x − 3 2 Y = y + 3 5 Z = z + 3 13 → ⎩ ⎨ ⎧ x = X + 3 2 y = Y − 3 5 z = Z − 3 13
Then,
5 x 2 + 4 y 2 − 4 y z + 2 x z + 2 x − 4 y − 8 z + 2 = = 5 ( X + 2 3 ) 2 + 4 ( Y − 5 3 ) 2 − 4 ( Y − 5 3 ) ( Z − 13 3 ) + 2 ( X + 2 3 ) ( Z − 13 3 ) + + 2 ( X + 2 3 ) − 4 ( Y − 5 3 ) − 8 ( Z − 13 3 ) + 2 = = 5 ( X 2 + 2 X ⋅ 2 3 + 4 9 ) + 4 ( Y 2 − 2 Y ⋅ 5 3 + 25 9 ) − 4 ( Y Z − 13 3 Y − 5 3 Z + 65 9 ) + + 2 ( X Z − 13 3 X + 2 3 Z − 26 9 ) + 2 X + 4 3 − 4 Y + 20 3 − 8 Z + 104 3 + 2 = = 5 X 2 + 20 3 X + 20 9 + 4 Y 2 − 40 3 Y + 100 9 − 4 Y Z + 52 3 Y + 20 3 Z − 260 9 + + 2 X Z − 26 3 X + 4 3 Z − 52 9 + 2 X + 4 3 − 4 Y + 20 3 − 8 Z + 104 3 + 2 = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + X ( 20 3 − 26 3 + 2 ) + Y ( − 40 3 + 52 3 − 4 ) + + Z ( 20 3 + 4 3 − 8 ) + ( 20 9 + 100 9 − 260 9 − 52 9 + 4 3 + 20 3 + 104 3 + 2 ) = \begin{array}{l}
5 x ^ {2} + 4 y ^ {2} - 4 y z + 2 x z + 2 x - 4 y - 8 z + 2 = \\
= 5 \left(X + \frac {2}{3}\right) ^ {2} + 4 \left(Y - \frac {5}{3}\right) ^ {2} - 4 \left(Y - \frac {5}{3}\right) \left(Z - \frac {13}{3}\right) + 2 \left(X + \frac {2}{3}\right) \left(Z - \frac {13}{3}\right) + \\
+ 2 \left(X + \frac {2}{3}\right) - 4 \left(Y - \frac {5}{3}\right) - 8 \left(Z - \frac {13}{3}\right) + 2 = \\
= 5 \left(X ^ {2} + 2 X \cdot \frac {2}{3} + \frac {4}{9}\right) + 4 \left(Y ^ {2} - 2 Y \cdot \frac {5}{3} + \frac {25}{9}\right) - 4 \left(Y Z - \frac {13}{3} Y - \frac {5}{3} Z + \frac {65}{9}\right) + \\
+ 2 \left(X Z - \frac {13}{3} X + \frac {2}{3} Z - \frac {26}{9}\right) + 2 X + \frac {4}{3} - 4 Y + \frac {20}{3} - 8 Z + \frac {104}{3} + 2 = \\
= 5 X ^ {2} + \frac {20}{3} X + \frac {20}{9} + 4 Y ^ {2} - \frac {40}{3} Y + \frac {100}{9} - 4 Y Z + \frac {52}{3} Y + \frac {20}{3} Z - \frac {260}{9} + \\
+ 2 X Z - \frac {26}{3} X + \frac {4}{3} Z - \frac {52}{9} + 2 X + \frac {4}{3} - 4 Y + \frac {20}{3} - 8 Z + \frac {104}{3} + 2 = \\
= 5 X ^ {2} + 4 Y ^ {2} - 4 Y Z + 2 X Z + X \left(\frac {20}{3} - \frac {26}{3} + 2\right) + Y \left(- \frac {40}{3} + \frac {52}{3} - 4\right) + \\
+ Z \left(\frac {20}{3} + \frac {4}{3} - 8\right) + \left(\frac {20}{9} + \frac {100}{9} - \frac {260}{9} - \frac {52}{9} + \frac {4}{3} + \frac {20}{3} + \frac {104}{3} + 2\right) = \\
\end{array} 5 x 2 + 4 y 2 − 4 yz + 2 x z + 2 x − 4 y − 8 z + 2 = = 5 ( X + 3 2 ) 2 + 4 ( Y − 3 5 ) 2 − 4 ( Y − 3 5 ) ( Z − 3 13 ) + 2 ( X + 3 2 ) ( Z − 3 13 ) + + 2 ( X + 3 2 ) − 4 ( Y − 3 5 ) − 8 ( Z − 3 13 ) + 2 = = 5 ( X 2 + 2 X ⋅ 3 2 + 9 4 ) + 4 ( Y 2 − 2 Y ⋅ 3 5 + 9 25 ) − 4 ( Y Z − 3 13 Y − 3 5 Z + 9 65 ) + + 2 ( XZ − 3 13 X + 3 2 Z − 9 26 ) + 2 X + 3 4 − 4 Y + 3 20 − 8 Z + 3 104 + 2 = = 5 X 2 + 3 20 X + 9 20 + 4 Y 2 − 3 40 Y + 9 100 − 4 Y Z + 3 52 Y + 3 20 Z − 9 260 + + 2 XZ − 3 26 X + 3 4 Z − 9 52 + 2 X + 3 4 − 4 Y + 3 20 − 8 Z + 3 104 + 2 = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + X ( 3 20 − 3 26 + 2 ) + Y ( − 3 40 + 3 52 − 4 ) + + Z ( 3 20 + 3 4 − 8 ) + ( 9 20 + 9 100 − 9 260 − 9 52 + 3 4 + 3 20 + 3 104 + 2 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + X ( 20 − 26 + 6 3 ) + Y ( − 40 + 52 − 12 3 ) + + Z ( 20 + 4 − 24 3 ) + ( 20 + 100 − 260 − 52 9 + 4 + 20 + 104 3 + 2 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + ( − 192 9 + 128 3 + 2 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + ( − 192 + 3 ⋅ 128 + 2 ⋅ 9 9 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + ( − 192 + 384 + 18 9 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + 210 9 = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + 70 3 \begin{array}{l}
= 5X^{2} + 4Y^{2} - 4YZ + 2XZ + X\left(\frac{20 - 26 + 6}{3}\right) + Y\left(\frac{-40 + 52 - 12}{3}\right) + \\
+ Z\left(\frac{20 + 4 - 24}{3}\right) + \left(\frac{20 + 100 - 260 - 52}{9} + \frac{4 + 20 + 104}{3} + 2\right) = \\
= 5X^{2} + 4Y^{2} - 4YZ + 2XZ + 0 \cdot X + 0 \cdot Y + 0 \cdot Z + \left(-\frac{192}{9} + \frac{128}{3} + 2\right) = \\
= 5X^{2} + 4Y^{2} - 4YZ + 2XZ + 0 \cdot X + 0 \cdot Y + 0 \cdot Z + \left(\frac{-192 + 3 \cdot 128 + 2 \cdot 9}{9}\right) = \\
= 5X^{2} + 4Y^{2} - 4YZ + 2XZ + 0 \cdot X + 0 \cdot Y + 0 \cdot Z + \left(\frac{-192 + 384 + 18}{9}\right) = \\
= 5X^{2} + 4Y^{2} - 4YZ + 2XZ + 0 \cdot X + 0 \cdot Y + 0 \cdot Z + \frac{210}{9} = \\
= 5X^{2} + 4Y^{2} - 4YZ + 2XZ + 0 \cdot X + 0 \cdot Y + 0 \cdot Z + \frac{70}{3}
\end{array} = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + X ( 3 20 − 26 + 6 ) + Y ( 3 − 40 + 52 − 12 ) + + Z ( 3 20 + 4 − 24 ) + ( 9 20 + 100 − 260 − 52 + 3 4 + 20 + 104 + 2 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + ( − 9 192 + 3 128 + 2 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + ( 9 − 192 + 3 ⋅ 128 + 2 ⋅ 9 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + ( 9 − 192 + 384 + 18 ) = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + 9 210 = = 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 0 ⋅ X + 0 ⋅ Y + 0 ⋅ Z + 3 70
Conclusion,
5 x 2 + 4 y 2 − 4 y z + 2 x z + 2 x − 4 y − 8 z + 2 = 0 → 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 70 3 = 0 5x^{2} + 4y^{2} - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 \rightarrow 5X^{2} + 4Y^{2} - 4YZ + 2XZ + \frac{70}{3} = 0 5 x 2 + 4 y 2 − 4 yz + 2 x z + 2 x − 4 y − 8 z + 2 = 0 → 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 3 70 = 0 ANSWER
1) The conicoid represented by
5 x 2 + 4 y 2 − 4 y z + 2 x z + 2 x − 4 y − 8 z + 2 = 0 5x^{2} + 4y^{2} - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 5 x 2 + 4 y 2 − 4 yz + 2 x z + 2 x − 4 y − 8 z + 2 = 0
isn't central.
2) The point P P P
P ( 2 3 , − 5 3 , − 13 3 ) is the center of the conicoid P\left(\frac{2}{3}, -\frac{5}{3}, -\frac{13}{3}\right) \text{ is the center of the conicoid} P ( 3 2 , − 3 5 , − 3 13 ) is the center of the conicoid
3) it is necessary to apply a shift of the form
{ x = X + 2 3 y = Y − 5 3 z = Z − 13 3 \left\{ \begin{array}{l} x = X + \frac {2}{3} \\ y = Y - \frac {5}{3} \\ z = Z - \frac {13}{3} \end{array} \right. ⎩ ⎨ ⎧ x = X + 3 2 y = Y − 3 5 z = Z − 3 13
to make the conicoid central.
5 x 2 + 4 y 2 − 4 y z + 2 x z + 2 x − 4 y − 8 z + 2 = 0 → 5 X 2 + 4 Y 2 − 4 Y Z + 2 X Z + 70 3 = 0 5 x ^ {2} + 4 y ^ {2} - 4 y z + 2 x z + 2 x - 4 y - 8 z + 2 = 0 \rightarrow 5 X ^ {2} + 4 Y ^ {2} - 4 Y Z + 2 X Z + \frac {70}{3} = 0 5 x 2 + 4 y 2 − 4 yz + 2 x z + 2 x − 4 y − 8 z + 2 = 0 → 5 X 2 + 4 Y 2 − 4 Y Z + 2 XZ + 3 70 = 0
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