Answer on Question #80359 – Math – Linear Algebra
Question
Reduce the conic x 2 + 6 x y + y 2 − 8 = 0 x^{2} + 6xy + y^{2} - 8 = 0 x 2 + 6 x y + y 2 − 8 = 0
Solution
The General Equation for a Conic Section
A x 2 + B x y + C y 2 + D x + E y + F = 0 Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 A x 2 + B x y + C y 2 + D x + E y + F = 0
In the given case we have x 2 + 6 x y + y 2 = 0 x^{2} + 6xy + y^{2} = 0 x 2 + 6 x y + y 2 = 0
A = 1 , B = 6 , C = 1 , D = 0 , E = 0 , F = − 8 A = 1, B = 6, C = 1, D = 0, E = 0, F = -8 A = 1 , B = 6 , C = 1 , D = 0 , E = 0 , F = − 8 B 2 − 4 A C = ( 36 ) 2 − 4 ( 1 ) ( 1 ) = 32 > 0 B^{2} - 4AC = (36)^{2} - 4(1)(1) = 32 > 0 B 2 − 4 A C = ( 36 ) 2 − 4 ( 1 ) ( 1 ) = 32 > 0
Then we have hyperbola or 2 intersecting lines.
A conic equation of the type of A x 2 + B x y + C y 2 + D x + E y + F = 0 Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 A x 2 + B x y + C y 2 + D x + E y + F = 0 θ \theta θ ( x ′ , y ′ ) (x', y') ( x ′ , y ′ ) θ \theta θ x y xy x y
The relation between coordinates ( x , y ) (x, y) ( x , y ) ( x ′ , y ′ ) (x', y') ( x ′ , y ′ )
x = x ′ cos θ − y ′ sin θ , y = x ′ sin θ + y ′ cos θ x = x' \cos \theta - y' \sin \theta, \quad y = x' \sin \theta + y' \cos \theta x = x ′ cos θ − y ′ sin θ , y = x ′ sin θ + y ′ cos θ
or
x ′ = x cos θ + y sin θ , y ′ = − x sin θ + y cos θ x' = x \cos \theta + y \sin \theta, \quad y' = -x \sin \theta + y \cos \theta x ′ = x cos θ + y sin θ , y ′ = − x sin θ + y cos θ
For this we need to have θ \theta θ
cot 2 θ = A − C B \cot 2\theta = \frac{A - C}{B} cot 2 θ = B A − C A = 1 , B = 6 , C = 1 A = 1, B = 6, C = 1 A = 1 , B = 6 , C = 1 cot 2 θ = 1 − 1 6 = 0 ⇒ θ = π 4 \cot 2\theta = \frac{1 - 1}{6} = 0 \Rightarrow \theta = \frac{\pi}{4} cot 2 θ = 6 1 − 1 = 0 ⇒ θ = 4 π
Then
x = x ′ cos π 4 − y ′ sin π 4 , y = x ′ sin π 4 + y ′ cos π 4 x = x' \cos \frac{\pi}{4} - y' \sin \frac{\pi}{4}, \quad y = x' \sin \frac{\pi}{4} + y' \cos \frac{\pi}{4} x = x ′ cos 4 π − y ′ sin 4 π , y = x ′ sin 4 π + y ′ cos 4 π x = x ′ 1 2 − y ′ 1 2 , y = x ′ 1 2 + y ′ 1 2 x = x' \frac{1}{\sqrt{2}} - y' \frac{1}{\sqrt{2}}, \quad y = x' \frac{1}{\sqrt{2}} + y' \frac{1}{\sqrt{2}} x = x ′ 2 1 − y ′ 2 1 , y = x ′ 2 1 + y ′ 2 1 x 2 + 6 x y + y 2 = 8 x^{2} + 6xy + y^{2} = 8 x 2 + 6 x y + y 2 = 8 ( x ′ 1 2 − y ′ 1 2 ) 2 + 6 ( x ′ 1 2 − y ′ 1 2 ) ( x ′ 1 2 + y ′ 1 2 ) + ( x ′ 1 2 + y ′ 1 2 ) 2 = 8 \left(x' \frac{1}{\sqrt{2}} - y' \frac{1}{\sqrt{2}}\right)^{2} + 6\left(x' \frac{1}{\sqrt{2}} - y' \frac{1}{\sqrt{2}}\right)\left(x' \frac{1}{\sqrt{2}} + y' \frac{1}{\sqrt{2}}\right) + \left(x' \frac{1}{\sqrt{2}} + y' \frac{1}{\sqrt{2}}\right)^{2} = 8 ( x ′ 2 1 − y ′ 2 1 ) 2 + 6 ( x ′ 2 1 − y ′ 2 1 ) ( x ′ 2 1 + y ′ 2 1 ) + ( x ′ 2 1 + y ′ 2 1 ) 2 = 8 1 2 x ′ 2 − x ′ y ′ + 1 2 y ′ 2 + 6 ( 1 2 ) x ′ 2 − 6 ( 1 2 ) y ′ 2 + 1 2 x ′ 2 + x ′ y ′ + 1 2 y ′ 2 = 8 \frac{1}{2} x'^{2} - x' y' + \frac{1}{2} y'^{2} + 6\left(\frac{1}{2}\right) x'^{2} - 6\left(\frac{1}{2}\right) y'^{2} + \frac{1}{2} x'^{2} + x' y' + \frac{1}{2} y'^{2} = 8 2 1 x ′ 2 − x ′ y ′ + 2 1 y ′ 2 + 6 ( 2 1 ) x ′ 2 − 6 ( 2 1 ) y ′ 2 + 2 1 x ′ 2 + x ′ y ′ + 2 1 y ′ 2 = 8 4 x ′ 2 − 2 y ′ 2 = 8 4x'^{2} - 2y'^{2} = 8 4 x ′ 2 − 2 y ′ 2 = 8 x ′ 2 2 − y ′ 2 4 = 1 \frac{x'^{2}}{2} - \frac{y'^{2}}{4} = 1 2 x ′ 2 − 4 y ′ 2 = 1
This is a canonical equation of hyperbola.
Or
x 2 + 6 x y + y 2 = 8 x ^ {2} + 6 x y + y ^ {2} = 8 x 2 + 6 x y + y 2 = 8 x 2 + 2 x ( 3 y ) + ( 3 y ) 2 − ( 3 y ) 2 + y 2 = 8 x ^ {2} + 2 x (3 y) + (3 y) ^ {2} - (3 y) ^ {2} + y ^ {2} = 8 x 2 + 2 x ( 3 y ) + ( 3 y ) 2 − ( 3 y ) 2 + y 2 = 8 ( x + 3 y ) 2 − 8 y 2 = 8 (x + 3 y) ^ {2} - 8 y ^ {2} = 8 ( x + 3 y ) 2 − 8 y 2 = 8
Substituting x ′ = x + 3 y , y ′ = y x' = x + 3y, y' = y x ′ = x + 3 y , y ′ = y
x ′ 2 − 8 y ′ 2 = 8 x ^ {\prime 2} - 8 y ^ {\prime 2} = 8 x ′2 − 8 y ′2 = 8 x ′ 2 8 − y ′ 2 1 = 1 \frac {x ^ {\prime 2}}{8} - \frac {y ^ {\prime 2}}{1} = 1 8 x ′2 − 1 y ′2 = 1
This is a canonical equation of hyperbola.
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