Equation and length of axes of x²−4xy−2y²+10x+4y=0

Here a = 1, h = −2, b = −2, so h² − ab = 4 + 2 = 6 > 0: a hyperbola.

Centre

x − 2y + 5 = 0 and −2x − 2y + 2 = 0. Solving: x = −1, y = 2. Centre: (−1, 2).

Shift the origin

Constant: g·x₀ + f·y₀ + c = 5(−1) + 2(2) + 0 = −1, so

X² − 4XY − 2Y² = 1

Principal directions

The roots of λ² − (a + b)λ + (ab − h²) = 0, i.e. λ² + λ − 6 = 0, are λ = 2 and λ = −3. The equation becomes

2u² − 3v² = 1, or u²/(1/2) − v²/(1/3) = 1

The direction of each axis is the eigenvector of the matrix [[1, −2], [−2, −2]]:

• λ = 2: (−2, 1) (check: first row gives 1·(−2) + (−2)(1) = −4 = 2·(−2)).
• λ = −3: (1, 2) (check: first row gives 1·1 + (−2)(2) = −3 = −3·1).

Equations of the axes (through the centre (−1, 2))

• Transverse axis (direction (−2, 1)): (x + 1)/(−2) = (y − 2)/1, i.e. x + 2y − 3 = 0.
• Conjugate axis (direction (1, 2)): (x + 1)/1 = (y − 2)/2, i.e. 2x − y + 4 = 0.

Lengths of the axes

• Transverse axis = 2 · (1/√2) = √2.
• Conjugate axis = 2 · (1/√3) = 2/√3.