Product of semi-axes of 5x²+6xy+5y²+12x+4y−4=0

Here a = 5, h = 3, b = 5, so h² − ab = 9 − 25 < 0: an ellipse.

Centre

5x + 3y + 6 = 0 and 3x + 5y + 2 = 0. Solving: x = −3/2, y = 1/2. Centre: (−3/2, 1/2).

Shift the origin

Constant: g·x₀ + f·y₀ + c = 6(−3/2) + 2(1/2) − 4 = −9 + 1 − 4 = −12, so

5X² + 6XY + 5Y² = 12

Rotate

Since a = b, rotate through 45°: X = (u − v)/√2, Y = (u + v)/√2. Then X² + Y² = u² + v² and XY = (u² − v²)/2:

5(u² + v²) + 3(u² − v²) = 12 8u² + 2v² = 12

u²/(3/2) + v²/6 = 1

Semi-axes

A = √(3/2), B = √6.

Product of the semi-axes = √(3/2 · 6) = √9 = 3. Hence proved.

(Check with the general result: product = |constant| / √(ab − h²) = 12/√16 = 3.)