---
title: "Remove linear terms by translation"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:translate-axes-3x2-2xy-4y2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/translate-axes-3x2-2xy-4y2
site: BCT Engineering
---

# Remove linear terms by translation

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2069 Asar
- Video: [Shifting of origin| Transformation of origin | rotation of origin | Removal of x and y terms.Bsc.](https://www.youtube.com/watch?v=7Bq_ExII0C0) — Instant learning
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/translate-axes-3x2-2xy-4y2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Transform the equation 3x² − 2xy + 4y² + 8x − 10y + 8 = 0 by translating the axes into an equation with the linear terms missing.

## Short answer

Find centre by ∂f/∂x=0, ∂f/∂y=0 → 3x−y+4=0, x−4y+5=0 → centre (−1,1). Shift origin: x=X−1, y=Y+1. Result: 3X²−2XY+4Y²=1.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. To remove the linear terms we shift the origin to?
   A. The vertex
   B. The focus
   C. Any point on the curve
   D. The centre of the conic
   Answer: D. Translating to the centre removes the first-degree terms. (Key point: Goal)

2. Which equations give the centre?
   A. f = 0 and x = y
   B. ∂f/∂x = ∂f/∂y
   C. ∂f/∂x = 0 and ∂f/∂y = 0, giving (−1, 1)
   D. h^2 = ab
   Answer: C. 3x − y + 4 = 0 and x − 4y + 5 = 0. (Key point: Centre)

3. What is the new constant?
   A. 8
   B. −1
   C. 0
   D. 1
   Answer: B. g·h + f·k + c = 4(−1) + (−5)(1) + 8 = −1. (Key point: Constant)

4. Transformed equation?
   A. 3X^2 − 2XY + 4Y^2 = 1
   B. 3X^2 − 2XY + 4Y^2 = 0
   C. 3X^2 + 4Y^2 = 1
   D. 3X^2 − 2XY + 4Y^2 = −1
   Answer: A. Quadratic part unchanged; constant −1 moves to the RHS. (Key point: Result)

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