Centre, foci, eccentricity of x²+4xy+y²−2x+2y−6=0

Here a = 1, 2h = 4 (h = 2), b = 1, so h² − ab = 3 > 0: the conic is a hyperbola.

Centre

Differentiating (or solving ax + hy + g = 0, hx + by + f = 0):

x + 2y − 1 = 0 and 2x + y + 1 = 0

Solving, x = −1, y = 1. Centre: (−1, 1).

Shift the origin to the centre

Put x = X − 1, y = Y + 1. The constant becomes g·x₀ + f·y₀ + c = (−1)(−1) + (1)(1) − 6 = −4, so

X² + 4XY + Y² = 4

Rotate the axes

Since a = b, the rotation angle is 45° (tan 2θ = 2h/(a − b) = ∞). Put X = (u − v)/√2, Y = (u + v)/√2. Then X² + Y² = u² + v² and XY = (u² − v²)/2, so

u² + v² + 2(u² − v²) = 4, i.e. 3u² − v² = 4

u²/(4/3) − v²/4 = 1

Here a² = 4/3, b² = 4.

Results

• Eccentricity: e² = 1 + b²/a² = 1 + 3 = 4, so e = 2.
• ae = √(a² + b²) = √(16/3) = 4/√3, so the foci are at u = ±4/√3, v = 0.
• Converting back (X = u/√2, Y = u/√2): X = Y = ±4/√6 = ±2√6/3.

Foci: (−1 + 2√6/3, 1 + 2√6/3) and (−1 − 2√6/3, 1 − 2√6/3).