Centre, axes, directrices of 3x²+8xy−3y²−40x−20y+50=0

Here a = 3, h = 4, b = −3, so h² − ab = 25 > 0 and a + b = 0: it is a rectangular hyperbola.

Centre

3x + 4y − 20 = 0 and 4x − 3y − 10 = 0. Solving: x = 4, y = 2. Centre: (4, 2).

Shift origin to the centre

Constant term: g·x₀ + f·y₀ + c = (−20)(4) + (−10)(2) + 50 = −50, so

3X² + 8XY − 3Y² = 50

Rotate

tan 2θ = 2h/(a − b) = 8/6 = 4/3, so cos 2θ = 3/5, tan θ = 1/2: cos θ = 2/√5, sin θ = 1/√5. The new coefficients are the roots of λ² − (a+b)λ + (ab − h²) = 0, i.e. λ² − 25 = 0, λ = ±5. Thus

5u² − 5v² = 50, or u²/10 − v²/10 = 1

where u is measured along the direction (2, 1)/√5.

Results

• a² = b² = 10, so the semi-axes are √10 each: length of axes = 2√10 (both).
• Eccentricity e = √(1 + b²/a²) = √2.
• Directrices in the new frame: u = ±a/e = ±√5. Since u = [2(x−4) + (y−2)]/√5 = (2x + y − 10)/√5,

2x + y − 10 = ±5

Directrices: 2x + y − 5 = 0 and 2x + y − 15 = 0.

• Axes: transverse x − 2y = 0; conjugate 2x + y − 10 = 0 (both through (4, 2)).
• Foci: u = ±ae = ±2√5, i.e. (4, 2) ± (4, 2) = (8, 4) and (0, 0).