---
title: "Centre, axes of 9x²+4xy+6y²−22x−16y+9=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:conic-9x2-4xy-6y2-centre-axes"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-9x2-4xy-6y2-centre-axes
site: BCT Engineering
---

# Centre, axes of 9x²+4xy+6y²−22x−16y+9=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2070 Asar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-9x2-4xy-6y2-centre-axes
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the centre, length of axes and eccentricity of the conic 9x² + 4xy + 6y² − 22x − 16y + 9 = 0.

## Short answer

Centre (1,1). Lengths of axes: major 2√2, minor 2. Eccentricity e = 1/√2.

## Answer

Here a = 9, h = 2, b = 6, so h² − ab = 4 − 54 < 0: an ellipse.

#### Centre
9x + 2y − 11 = 0 and 2x + 6y − 8 = 0. Solving: x = 1, y = 1. **Centre: (1, 1).**

#### Shift the origin
Constant: g·x₀ + f·y₀ + c = (−11)(1) + (−8)(1) + 9 = −10, so

9X² + 4XY + 6Y² = 10

#### Remove XY by rotation
The new coefficients are the roots of λ² − 15λ + (54 − 4) = 0, i.e. λ² − 15λ + 50 = 0, so λ = 5 and 10. Hence

5u² + 10v² = 10, or **u²/2 + v²/1 = 1**

#### Results
- Semi-axes: √2 and 1. **Lengths of axes: 2√2 (major) and 2 (minor).**
- **Eccentricity:** e² = 1 − b²/a² = 1 − 1/2 = 1/2, so **e = 1/√2**.
- The major axis is along the eigen-direction of λ = 5, i.e. (−1, 2): the line 2x + y − 3 = 0; the minor axis is x − 2y + 1 = 0.

## Check yourself

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

1. What is the centre of 9x^2 + 4xy + 6y^2 − 22x − 16y + 9 = 0?
   A. (2, 1)
   B. (1, 2)
   C. (1, 1)
   D. (11/9, 8/6)
   Answer: C. 9x + 2y − 11 = 0 and 2x + 6y − 8 = 0 give (1, 1). (Key point: Centre)

2. After shifting, what is the right-hand side?
   A. 9X^2 + 4XY + 6Y^2 = 9
   B. 9X^2 + 4XY + 6Y^2 = 10
   C. 9X^2 + 4XY + 6Y^2 = −10
   D. 9X^2 + 4XY + 6Y^2 = 0
   Answer: B. Constant g·x0 + f·y0 + c = −11 − 8 + 9 = −10, so the RHS is +10. (Key point: Shifted equation)

3. What are the new coefficients?
   A. 5 and 10 (roots of λ^2 − 15λ + 50 = 0)
   B. 9 and 6
   C. 3 and 12
   D. 15 and 50
   Answer: A. Sum = 15, product = 54 − 4 = 50. (Key point: Rotation)

4. From u^2/2 + v^2/1 = 1, what are axes and e?
   A. sqrt2 and 1; e = 1/sqrt2
   B. 2sqrt2 and 2; e = 1/2
   C. 2 and 1; e = 1/2
   D. 2sqrt2 and 2; e = 1/sqrt2
   Answer: D. Semi-axes sqrt2 and 1; e^2 = 1 − 1/2 = 1/2. (Key point: Axes and eccentricity)

## Close variants set in other exams

- [Centre, axes, directrices of 3x²+8xy−3y²−40x−20y+50=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-3x2-8xy-3y2-directrices) — Mathematics I, chapter 3, asked 7×
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