Remove linear terms by translation

To remove the linear terms, shift the origin to the centre of the conic.

Find the centre (h, k)

The centre satisfies ∂f/∂x = 0 and ∂f/∂y = 0 for f = 3x² − 2xy + 4y² + 8x − 10y + 8:

6x − 2y + 8 = 0 → 3x − y + 4 = 0 −2x + 8y − 10 = 0 → x − 4y + 5 = 0

From the second, x = 4y − 5. Then 12y − 15 − y + 4 = 0, so y = 1 and x = −1. Centre (−1, 1).

Translate

Put x = X − 1, y = Y + 1:

• 3x² = 3(X² − 2X + 1) = 3X² − 6X + 3
• −2xy = −2(X − 1)(Y + 1) = −2XY − 2X + 2Y + 2
• 4y² = 4(Y² + 2Y + 1) = 4Y² + 8Y + 4
• 8x = 8X − 8
• −10y = −10Y − 10
• +8

Sum: X-terms: −6X − 2X + 8X = 0. Y-terms: 2Y + 8Y − 10Y = 0. Constants: 3 + 2 + 4 − 8 − 10 + 8 = −1.

Transformed equation: 3X² − 2XY + 4Y² − 1 = 0, i.e. 3X² − 2XY + 4Y² = 1.

The constant can also be found as g·h + f·k + c = 4(−1) + (−5)(1) + 8 = −1.