Hyperbola : find the equation of the principal axes of the hyperbola 2x^2- 3xy - 2y^2 + 3x - y +8 = 0? explain?

Ax² + Bxy + Cy² + Dx + Ey + F = 0

A = 2

B = -3

C = -2

D = 3

E = -1

F = 8

A rotation by angle θ would turn the coordinate axes into a position square with the axes of the hyperbola, (or any other conic).

cot(2θ) = (A - C)/B

cot(2θ) = -4/3

cos(2θ) = -4/5

sinθ = √[(1 - cos(2θ))/2] = 3/√(10)

cosθ = √[(1 + cos(2θ))/2] = 1/√(10)

Rotate the coordinate system by angle θ. In the new system, the hyperbola has parameters A'. B', ...

A' = Acos²θ + Bsinθcosθ + Csin²θ = -5/2

B' = B(cos²θ - sin²θ) - 2(A - C)sinθcosθ = 0

C' = Asin²θ - Bsinθcosθ + Ccos²θ = 5/2

D' = Dcosθ + Esinθ = 0

E' = -Dsinθ + Ecosθ = -√(10)

F' = F = 8

-(5/2)x'² + (5/2)y'² - √(10)y + 8 = 0

In the x'-y' system, it is a rectangular hyperbola with center (0, √(10)/5). Rotate the center to the x-y system.

x = (0)cosθ - [√(10)/5]sinθ = -3/5

x = (0)sinθ + [√(10)/5]cosθ = 1/5

center: (-3/5, 1/5)

The slope of the transverse axis is tanθ = 3, and the other axis has slope -1/3.

y - 1/5 = 3(x + 3/5)

3x - y + 2 = 0

y - 1/5 = (-1/3)(x + 3/5)

15y - 3 = -5x - 3

x + 3y = 0

2x^2 - 3xy - 2y^2 + 3x - y + 8 = 0 is a rotated conic. After a rotation of about 18°, the new equation is 2.5x'^2 - 2.5y'^2 + 3.16227766...x' + 8 = 0.

(5/2)x'^2 - (5/2)y'^2 + √10x' = -8

(5/16)y'^2 - (5/16)x'^2 - 8√10x' = 1

y'^2/(16/5) - x'^2/(16/5) - 8√10x' = 1

y'^2/(16/5) - x'^2/(16/5) - (5/2)√10x' = 1

y'^2/(16/5) - (5/16)(x'^2 + 8√10x' + 160) = 1 - 50

y'^2/(16/5) - (x' - 4√10)^2/(16/5) = -49

(x' - 4√10)^2/(16/245) - y'^2/(16/245) = 1

The center is (4√10, 0) in the rotated axes, but I don't know what the coordinates are in the usual axes.

I'm not sure about the transverse or conjugate axes.

Did you try Wolfram Alpha?

http://www.wolframalpha.com/input/?i=2x^2-+3xy+-+2...