---
title: "Equation and length of axes of x²−4xy−2y²+10x+4y=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:conic-x2-4xy-2y2-equation-axes"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-x2-4xy-2y2-equation-axes
site: BCT Engineering
---

# Equation and length of axes of x²−4xy−2y²+10x+4y=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2068 Chaitra
- Video: [Solid Geometry - Length of Axes, Equation of Axes - Eccentricity of a Conic - BSC 1st Year Chapter 1](https://www.youtube.com/watch?v=qfj_njg8vV0) — Dr. Vikas Poply - MATHS EASY
- 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-2y2-equation-axes
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the equation of axes and length of axes of the conic x² − 4xy − 2y² + 10x + 4y = 0.

## Short answer

Transverse axis: x+2y−3=0, length √2. Conjugate axis: 2x−y+4=0, length 2/√3.

## Answer

Here a = 1, h = −2, b = −2, so h² − ab = 4 + 2 = 6 > 0: a hyperbola.

#### Centre
x − 2y + 5 = 0 and −2x − 2y + 2 = 0. Solving: x = −1, y = 2. **Centre: (−1, 2).**

#### Shift the origin
Constant: g·x₀ + f·y₀ + c = 5(−1) + 2(2) + 0 = −1, so

X² − 4XY − 2Y² = 1

#### Principal directions
The roots of λ² − (a + b)λ + (ab − h²) = 0, i.e. λ² + λ − 6 = 0, are λ = 2 and λ = −3. The equation becomes

2u² − 3v² = 1, or **u²/(1/2) − v²/(1/3) = 1**

The direction of each axis is the eigenvector of the matrix [[1, −2], [−2, −2]]:
- λ = 2: (−2, 1) (check: first row gives 1·(−2) + (−2)(1) = −4 = 2·(−2)).
- λ = −3: (1, 2) (check: first row gives 1·1 + (−2)(2) = −3 = −3·1).

#### Equations of the axes (through the centre (−1, 2))
- Transverse axis (direction (−2, 1)): (x + 1)/(−2) = (y − 2)/1, i.e. **x + 2y − 3 = 0**.
- Conjugate axis (direction (1, 2)): (x + 1)/1 = (y − 2)/2, i.e. **2x − y + 4 = 0**.

#### Lengths of the axes
- Transverse axis = 2 · (1/√2) = **√2**.
- Conjugate axis = 2 · (1/√3) = **2/√3**.

## Check yourself

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

1. What are the λ values for a=1,h=−2,b=−2?
   A. 1 and −2
   B. 3 and −2
   C. 2 and −3 (roots of λ^2 + λ − 6 = 0)
   D. 6 and −1
   Answer: C. Sum a + b = −1, product ab − h^2 = −6. (Key point: Principal values)

2. How are the directions of axes found?
   A. By solving h = 0
   B. As eigenvectors of [[1,−2],[−2,−2]] for each λ
   C. By the slope of the asymptotes only
   D. By substituting x = y
   Answer: B. λ = 2 gives (−2,1) (transverse), λ = −3 gives (1,2) (conjugate). (Key point: Axis directions)

3. What are the axes through the centre (−1, 2)?
   A. x + 2y − 3 = 0 and 2x − y + 4 = 0
   B. 2x + y = 0 and x − 2y = 0
   C. x − 2y + 5 = 0 and 2x + y = 0
   D. x + 2y = 0 and 2x − y = 0
   Answer: A. Lines through (−1,2) with directions (−2,1) and (1,2). (Key point: Equations of axes)

4. What are the lengths of the axes from 2u^2 − 3v^2 = 1?
   A. Transverse 2sqrt2; conjugate 2sqrt3
   B. Transverse 1/sqrt2; conjugate 1/sqrt3
   C. Transverse 2; conjugate 3
   D. Transverse sqrt2; conjugate 2/sqrt3
   Answer: D. a^2 = 1/2, b^2 = 1/3; lengths 2a = sqrt2 and 2b = 2/sqrt3. (Key point: Lengths)

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