---
title: "Centre, foci, eccentricity of x²+4xy+y²−2x+2y−6=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:conic-x2-4xy-y2-foci-eccentricity"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-x2-4xy-y2-foci-eccentricity
site: BCT Engineering
---

# Centre, foci, eccentricity of x²+4xy+y²−2x+2y−6=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2074 Asoj; 2073 Chaitra; 2072 Kartik (asked 3 times)
- Video: [Canonical form and the nature of the conic 7x^2-6xy-y^2+4x-4y-2=0](https://www.youtube.com/watch?v=nWQlF9waY24) — MATHS A TO Z
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-x2-4xy-y2-foci-eccentricity
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the centre, foci and eccentricity of the conic x² + 4xy + y² − 2x + 2y − 6 = 0.

## Short answer

Centre (−1,1); after rotation, standard form u²/(4/3)−v²/4=1. Eccentricity e=2; foci at (−1±2√6/3, 1±2√6/3).

## Answer

Here a = 1, 2h = 4 (h = 2), b = 1, so h² − ab = 3 > 0: the conic is a hyperbola.

#### Centre
Differentiating (or solving ax + hy + g = 0, hx + by + f = 0):

x + 2y − 1 = 0 and 2x + y + 1 = 0

Solving, x = −1, y = 1. **Centre: (−1, 1).**

#### Shift the origin to the centre
Put x = X − 1, y = Y + 1. The constant becomes g·x₀ + f·y₀ + c = (−1)(−1) + (1)(1) − 6 = −4, so

X² + 4XY + Y² = 4

#### Rotate the axes
Since a = b, the rotation angle is 45° (tan 2θ = 2h/(a − b) = ∞). Put X = (u − v)/√2, Y = (u + v)/√2. Then X² + Y² = u² + v² and XY = (u² − v²)/2, so

u² + v² + 2(u² − v²) = 4, i.e. 3u² − v² = 4

**u²/(4/3) − v²/4 = 1**

Here a² = 4/3, b² = 4.

#### Results
- **Eccentricity:** e² = 1 + b²/a² = 1 + 3 = 4, so **e = 2**.
- ae = √(a² + b²) = √(16/3) = 4/√3, so the foci are at u = ±4/√3, v = 0.
- Converting back (X = u/√2, Y = u/√2): X = Y = ±4/√6 = ±2√6/3.

**Foci: (−1 + 2√6/3, 1 + 2√6/3) and (−1 − 2√6/3, 1 − 2√6/3).**

## Check yourself

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

1. For x^2 + 4xy + y^2 − 2x + 2y − 6 = 0 (a=1,h=2,b=1) what type is the conic?
   A. Ellipse, since h^2 − ab > 0
   B. Hyperbola, since h^2 − ab = 3 > 0
   C. Parabola, since a = b
   D. Circle, since a = b
   Answer: B. h^2 − ab > 0 indicates a hyperbola. (Key point: Classification)

2. How is the centre found?
   A. Solve ax + hy + g = 0 and hx + by + f = 0, giving (−1, 1)
   B. Set x = y = 0
   C. Solve h^2 − ab = 0
   D. Complete the square on x only
   Answer: A. The centre is where both partial derivatives vanish: x + 2y − 1 = 0 and 2x + y + 1 = 0 give (−1, 1). (Key point: Finding the centre)

3. Why is the axes rotation 45°?
   A. Because h = 2
   B. Because the centre is (−1, 1)
   C. Because the conic is a parabola
   D. Since a = b, tan2θ = 2h/(a − b) is infinite so 2θ = 90°
   Answer: D. With a = b the denominator is zero, so 2θ = 90°, θ = 45°. (Key point: Rotation angle)

4. After rotation, 3u^2 − v^2 = 4 i.e. u^2/(4/3) − v^2/4 = 1. What is e?
   A. sqrt2
   B. 1/2
   C. 2
   D. sqrt3
   Answer: C. e^2 = 1 + b^2/a^2 = 1 + 4/(4/3) = 4, so e = 2. (Key point: Eccentricity)

## Related in Mathematics I

- [Axes of conic 14x²−4xy+11y²−44x−58y+71=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-14x2-4xy-11y2-axes) — Mathematics I, chapter 3, asked 2×
- [Centre, axes of 9x²+4xy+6y²−22x−16y+9=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-9x2-4xy-6y2-centre-axes) — Mathematics I, chapter 3, asked 7×
- [Equation and length of axes of x²−4xy−2y²+10x+4y=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-x2-4xy-2y2-equation-axes) — Mathematics I, chapter 3
- [Centre, axes of ellipse 2x²+3y²−4x−12y+13=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aellipse-2x2-3y2-centre-axes) — Mathematics I, chapter 3, asked 7×
- [Solve y''+3y'+2y=e^(2x) sin x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-d2-3d-2-e2x-sinx) — Mathematics I, chapter 4, asked 2×
- [Solve dy/dx = (x+2y−3)/(2x+y−3)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-homogeneous-linear-x-2y-3) — Mathematics I, chapter 4