Answer on Question #62049 – Math – Linear Algebra
Question
Find the nature of the conic
2 x 2 + x y − y 2 − 6 = 0. 2x^2 + xy - y^2 - 6 = 0. 2 x 2 + x y − y 2 − 6 = 0.
Solution
The general equation of the second degree in two variables is
A x 2 + 2 B x y + C y 2 + 2 D x + 2 E y + F = 0 A = 2 ; 2 B = 1 ; C = − 1 ; 2 D = 0 ; 2 E = 0 ; F = − 6. \begin{array}{l}
Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0 \\
A = 2; 2B = 1; C = -1; 2D = 0; 2E = 0; F = -6.
\end{array} A x 2 + 2 B x y + C y 2 + 2 D x + 2 E y + F = 0 A = 2 ; 2 B = 1 ; C = − 1 ; 2 D = 0 ; 2 E = 0 ; F = − 6.
a) Simplification by rotation of coordinate system.
With a suitable rotation of the coordinate system the term in x y xy x y θ \theta θ
tan 2 θ = 2 B A − C = 1 2 − ( − 1 ) = 1 3 . \tan 2\theta = \frac{2B}{A - C} = \frac{1}{2 - (-1)} = \frac{1}{3}. tan 2 θ = A − C 2 B = 2 − ( − 1 ) 1 = 3 1 . x = x ′ cos θ − y ′ sin θ ; y = x ′ sin θ + y ′ cos θ ; \begin{array}{l}
x = x' \cos \theta - y' \sin \theta; \\
y = x' \sin \theta + y' \cos \theta;
\end{array} x = x ′ cos θ − y ′ sin θ ; y = x ′ sin θ + y ′ cos θ ; cos 2 θ = 1 1 + ( tan 2 θ ) 2 = 3 10 ; sin θ = 1 − cos 2 θ 2 = 10 − 3 2 10 ; cos θ = 1 + cos 2 θ 2 = 10 + 3 2 10 . x = x ′ 10 + 3 2 10 − y ′ 10 − 3 2 10 ; y = x ′ 10 − 3 2 10 + y ′ 10 + 3 2 10 ; \begin{array}{l}
\cos 2\theta = \frac{1}{\sqrt{1 + (\tan 2\theta)^2}} = \frac{3}{\sqrt{10}}; \\
\sin \theta = \sqrt{\frac{1 - \cos 2\theta}{2}} = \sqrt{\frac{\sqrt{10} - 3}{2\sqrt{10}}}; \\
\cos \theta = \sqrt{\frac{1 + \cos 2\theta}{2}} = \sqrt{\frac{\sqrt{10} + 3}{2\sqrt{10}}}. \\
x = x' \sqrt{\frac{\sqrt{10} + 3}{2\sqrt{10}}} - y' \sqrt{\frac{\sqrt{10} - 3}{2\sqrt{10}}}; \\
y = x' \sqrt{\frac{\sqrt{10} - 3}{2\sqrt{10}}} + y' \sqrt{\frac{\sqrt{10} + 3}{2\sqrt{10}}};
\end{array} cos 2 θ = 1 + ( t a n 2 θ ) 2 1 = 10 3 ; sin θ = 2 1 − c o s 2 θ = 2 10 10 − 3 ; cos θ = 2 1 + c o s 2 θ = 2 10 10 + 3 . x = x ′ 2 10 10 + 3 − y ′ 2 10 10 − 3 ; y = x ′ 2 10 10 − 3 + y ′ 2 10 10 + 3 ;
the general equation of the second degree becomes
A ′ x ′ 2 + 2 B ′ x ′ y ′ + C ′ y ′ 2 + 2 D ′ x ′ + 2 E ′ y ′ + F ′ = 0 A'x'^2 + 2B'x'y' + C'y'^2 + 2D'x' + 2E'y' + F' = 0 A ′ x ′2 + 2 B ′ x ′ y ′ + C ′ y ′2 + 2 D ′ x ′ + 2 E ′ y ′ + F ′ = 0
where
A ′ = A ( cos θ ) 2 + 2 B sin θ cos θ + C ( sin θ ) 2 B ′ = ( C − A ) sin θ cos θ + B ( ( cos θ ) 2 − ( sin θ ) 2 ) = 1 2 ( C − A ) sin 2 θ + B cos 2 θ C ′ = A ( sin θ ) 2 − 2 B sin θ cos θ + C ( cos θ ) 2 D ′ = D cos θ + E sin θ E ′ = E cos θ − D sin θ \begin{array}{l}
A' = A(\cos \theta)^2 + 2B \sin \theta \cos \theta + C(\sin \theta)^2 \\
B' = (C - A) \sin \theta \cos \theta + B((\cos \theta)^2 - (\sin \theta)^2) = \frac{1}{2} (C - A) \sin 2\theta + B \cos 2\theta \\
C' = A(\sin \theta)^2 - 2B \sin \theta \cos \theta + C(\cos \theta)^2 \\
D' = D \cos \theta + E \sin \theta \\
E' = E \cos \theta - D \sin \theta \\
\end{array} A ′ = A ( cos θ ) 2 + 2 B sin θ cos θ + C ( sin θ ) 2 B ′ = ( C − A ) sin θ cos θ + B (( cos θ ) 2 − ( sin θ ) 2 ) = 2 1 ( C − A ) sin 2 θ + B cos 2 θ C ′ = A ( sin θ ) 2 − 2 B sin θ cos θ + C ( cos θ ) 2 D ′ = D cos θ + E sin θ E ′ = E cos θ − D sin θ F ′ = F A ′ = 2 ( 10 + 3 2 10 ) + ( 10 − 3 2 10 ) ( 10 + 3 2 10 ) − ( 10 − 3 2 10 ) = 1 + 10 2 B ′ = ( − 1 − 2 ) ( 10 − 3 2 10 ) ( 10 + 3 2 10 ) + 1 2 ( 10 + 3 2 10 − 10 − 3 2 10 ) = 0 C ′ = 2 ( 10 − 3 2 10 ) + ( 10 − 3 2 10 ) ( 10 + 3 2 10 ) − ( 10 + 3 2 10 ) = 1 − 10 2 D ′ = 0 E ′ = 0 F ′ = − 6 1 + 10 2 x ′ 2 + 1 − 10 2 y ′ 2 − 6 = 0 x ′ 2 [ 12 ⋅ ( 10 − 1 ) ] 2 − y ′ 2 [ 12 ⋅ ( 10 + 1 ) ] 2 = 1 \begin{array}{l}
F' = F \\
A' = 2 \left( \frac{\sqrt{10} + 3}{2\sqrt{10}} \right) + \sqrt{ \left( \frac{\sqrt{10} - 3}{2\sqrt{10}} \right) \left( \frac{\sqrt{10} + 3}{2\sqrt{10}} \right) } - \left( \frac{\sqrt{10} - 3}{2\sqrt{10}} \right) = \frac{1 + \sqrt{10}}{2} \\
B' = (-1 - 2) \sqrt{ \left( \frac{\sqrt{10} - 3}{2\sqrt{10}} \right) \left( \frac{\sqrt{10} + 3}{2\sqrt{10}} \right) } + \frac{1}{2} \left( \frac{\sqrt{10} + 3}{2\sqrt{10}} - \frac{\sqrt{10} - 3}{2\sqrt{10}} \right) = 0 \\
C' = 2 \left( \frac{\sqrt{10} - 3}{2\sqrt{10}} \right) + \sqrt{ \left( \frac{\sqrt{10} - 3}{2\sqrt{10}} \right) \left( \frac{\sqrt{10} + 3}{2\sqrt{10}} \right) } - \left( \frac{\sqrt{10} + 3}{2\sqrt{10}} \right) = \frac{1 - \sqrt{10}}{2} \\
D' = 0 \\
E' = 0 \\
F' = -6 \\
\frac{ \frac{1 + \sqrt{10}}{2} x'^2 + \frac{1 - \sqrt{10}}{2} y'^2 - 6 = 0 }{ \frac{x'^2}{[\sqrt{12 \cdot (\sqrt{10} - 1)}]^2} - \frac{y'^2}{[\sqrt{12 \cdot (\sqrt{10} + 1)}]^2} } = 1 \\
\end{array} F ′ = F A ′ = 2 ( 2 10 10 + 3 ) + ( 2 10 10 − 3 ) ( 2 10 10 + 3 ) − ( 2 10 10 − 3 ) = 2 1 + 10 B ′ = ( − 1 − 2 ) ( 2 10 10 − 3 ) ( 2 10 10 + 3 ) + 2 1 ( 2 10 10 + 3 − 2 10 10 − 3 ) = 0 C ′ = 2 ( 2 10 10 − 3 ) + ( 2 10 10 − 3 ) ( 2 10 10 + 3 ) − ( 2 10 10 + 3 ) = 2 1 − 10 D ′ = 0 E ′ = 0 F ′ = − 6 [ 12 ⋅ ( 10 − 1 ) ] 2 x ′2 − [ 12 ⋅ ( 10 + 1 ) ] 2 y ′2 2 1 + 10 x ′2 + 2 1 − 10 y ′2 − 6 = 0 = 1
This is the equation of hyperbola:
a = 12 ⋅ ( 10 − 1 ) ; b = 12 ⋅ ( 10 + 1 ) c 2 = a 2 + b 2 = 12 ⋅ ( 10 − 1 + 10 + 1 ) = 24 ⋅ 10 . \begin{array}{l}
a = \sqrt{12 \cdot (\sqrt{10} - 1)}; \; b = \sqrt{12 \cdot (\sqrt{10} + 1)} \; c^2 = a^2 + b^2 \\
= 12 \cdot (\sqrt{10} - 1 + \sqrt{10} + 1) = 24 \cdot \sqrt{10}.
\end{array} a = 12 ⋅ ( 10 − 1 ) ; b = 12 ⋅ ( 10 + 1 ) c 2 = a 2 + b 2 = 12 ⋅ ( 10 − 1 + 10 + 1 ) = 24 ⋅ 10 .
Answer: hyperbola.
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