---
title: "Axes of conic 14x²−4xy+11y²−44x−58y+71=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:conic-14x2-4xy-11y2-axes"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-14x2-4xy-11y2-axes
site: BCT Engineering
---

# Axes of conic 14x²−4xy+11y²−44x−58y+71=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2071 Shrawan, 5 marks; 2071 Chaitra (asked 2 times)
- Video: [Reduce the Second degree equation into standard form/redduction of second degree equation bsc1year](https://www.youtube.com/watch?v=b5PQICAV0Wk) — Bê Unique
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-14x2-4xy-11y2-axes
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the length of axes and eccentricity of the conic 14x² − 4xy + 11y² − 44x − 58y + 71 = 0. (Also show that it is an ellipse and find its centre, eccentricity, latus rectum and foci.)

## Short answer

Complete squares and rotate: ellipse with centre (2,3). Standard form u²/6+v²/4=1. Axes lengths 2√6 and 4; eccentricity 1/√3. Latus rectum 4√6/3; foci (2±√10/5, 3±2√10/5).

## Answer

Here a = 14, h = −2, b = 11, so h² − ab = 4 − 154 < 0: the conic is an **ellipse** (and Δ ≠ 0, so it is not degenerate).

#### Centre
14x − 2y − 22 = 0 and −2x + 11y − 29 = 0. Solving: x = 2, y = 3. **Centre: (2, 3).**

#### Shift origin
Constant: g·x₀ + f·y₀ + c = (−22)(2) + (−29)(3) + 71 = −60, so

14X² − 4XY + 11Y² = 60

#### Rotate
The coefficients of the squares after rotation are the roots of λ² − (a+b)λ + (ab − h²) = 0:

λ² − 25λ + 150 = 0, so λ = 10 and 15.

The new equation is 10u² + 15v² = 60, i.e.

**u²/6 + v²/4 = 1**

where u is along the eigenvector of λ = 10, that is the direction (1, 2)/√5, and v along (−2, 1)/√5.

#### Results
- Semi-axes: a = √6 (major), b = 2 (minor). **Lengths of axes: 2√6 and 4.**
- **Eccentricity:** e² = 1 − b²/a² = 1 − 4/6 = 1/3, so e = 1/√3.
- **Latus rectum:** 2b²/a = 8/√6 = 4√6/3.
- **Foci:** ae = √(a² − b²) = √2 from the centre along (1, 2)/√5, giving
  (2 ± √10/5, 3 ± 2√10/5).
- Major axis line: 2x − y − 1 = 0; minor axis line: x + 2y − 8 = 0.

## Check yourself

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

1. Why is 14x^2 − 4xy + 11y^2 − ... an ellipse?
   A. h^2 − ab > 0
   B. h^2 − ab = 4 − 154 < 0 and the conic is non-degenerate
   C. a ≠ b
   D. Because the constant is positive
   Answer: B. h^2 − ab < 0 with Δ ≠ 0 means a genuine ellipse. (Key point: Classification)

2. What is the centre?
   A. (2, 3)
   B. (3, 2)
   C. (−2, −3)
   D. (1, 1)
   Answer: A. 14x − 2y − 22 = 0 and −2x + 11y − 29 = 0 give x = 2, y = 3. (Key point: Centre)

3. What are the new coefficients of u^2, v^2?
   A. 14 and 11
   B. 25 and 150
   C. 5 and 30
   D. 10 and 15, the roots of λ^2 − 25λ + 150 = 0
   Answer: D. Sum 25 and product ab − h^2 = 150, so λ = 10, 15. (Key point: Rotation coefficients)

4. Given u^2/6 + v^2/4 = 1, what are axis lengths and e?
   A. sqrt6 and 2; e = 1/sqrt3
   B. 2sqrt6 and 4; e = 2/3
   C. 2sqrt6 and 4; e = 1/sqrt3
   D. 6 and 4; e = 1/3
   Answer: C. A = sqrt6, B = 2; lengths 2A and 2B; e^2 = 1 − 4/6 = 1/3. (Key point: Axes and eccentricity)

## Related in Mathematics I

- [Centre, foci, eccentricity of x²+4xy+y²−2x+2y−6=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-x2-4xy-y2-foci-eccentricity) — Mathematics I, chapter 3, asked 3×
- [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×
- [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×
- [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, 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×
- [Product of semi-axes of x²−4xy+5y²=2](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-x2-4xy-5y2-product-semi-axes) — Mathematics I, chapter 3