---
title: "Centre, axes of ellipse 2x²+3y²−4x−12y+13=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:ellipse-2x2-3y2-centre-axes"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/ellipse-2x2-3y2-centre-axes
site: BCT Engineering
---

# Centre, axes of ellipse 2x²+3y²−4x−12y+13=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2072 Chaitra; 2071 Shrawan, 5 marks; 2070 Chaitra; 2067 Asar (asked 4 times)
- Video: [Ellipse | Definition, Foci, Vertex, Axes, Latus Rectum of Ellipse | 2D GEOMETRY - By GP Sir](https://www.youtube.com/watch?v=HVs4bNR4hxE) — Dr.Gajendra Purohit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/ellipse-2x2-3y2-centre-axes
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the centre, length of axes, eccentricity and directrices of the conic (ellipse) 2x² + 3y² − 4x − 12y + 13 = 0.

## Short answer

Complete squares: 2(x-1)²+3(y-2)²=1. Centre (1,2); major axis length √2, minor 2/√3; eccentricity 1/√3; directrices x=1±√6/2.

## Answer

Complete the squares to bring the conic to the standard form (x−h)²/a² + (y−k)²/b² = 1.

#### Reduction
2x² − 4x + 3y² − 12y + 13 = 0

2(x² − 2x) + 3(y² − 4y) = −13

2(x−1)² + 3(y−2)² = −13 + 2 + 12 = 1

Shift the origin to (1, 2): X = x − 1, Y = y − 2. Then

2X² + 3Y² = 1, that is X²/(1/2) + Y²/(1/3) = 1

So a² = 1/2 (major, since 1/2 > 1/3) and b² = 1/3.

#### Results
- **Centre:** (1, 2).
- **Axes:** major axis 2a = 2/√2 = √2 (parallel to the x-axis); minor axis 2b = 2/√3.
- **Eccentricity:** e² = 1 − b²/a² = 1 − (1/3)/(1/2) = 1/3, so **e = 1/√3**.
- **Foci:** ae = √(1/2) · (1/√3) = 1/√6, so the foci are (1 ± 1/√6, 2).
- **Directrices:** X = ±a/e = ±(1/√2)·√3 = ±√(3/2), that is x − 1 = ±√6/2, so

**x = 1 ± √6/2**

Check: a² − b² = 1/2 − 1/3 = 1/6 = (ae)², which matches ae = 1/√6.

## Check yourself

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

1. How do we reduce 2x^2 + 3y^2 − 4x − 12y + 13 = 0 to standard form?
   A. Rotate the axes through 45°
   B. Complete the squares in x and y, then shift the origin to the centre
   C. Differentiate and set equal to zero only
   D. Substitute y = mx
   Answer: B. No xy term is present, so completing the squares and translating the origin is enough. (Key point: Method)

2. What is the centre of the conic?
   A. (1, 2)
   B. (2, 1)
   C. (−1, −2)
   D. (1, 3)
   Answer: A. 2(x−1)^2 + 3(y−2)^2 = 1, so the centre is (1, 2). (Key point: Centre)

3. In X^2/(1/2) + Y^2/(1/3) = 1 which is the major axis and its length?
   A. b^2 = 1/3 gives major axis 2/sqrt3
   B. Major axis = 2 along y
   C. Major axis = 1/2
   D. a^2 = 1/2 is larger, so major axis = sqrt2 along the x-direction
   Answer: D. The larger denominator 1/2 belongs to X, so a = 1/sqrt2 and 2a = sqrt2. (Key point: Major axis and length)

4. What are e and the directrices?
   A. e = 1/3; x = 1 ± sqrt6
   B. e = sqrt3; x = 1 ± 1/sqrt6
   C. e = 1/sqrt3; x = 1 ± sqrt6/2
   D. e = 1/sqrt2; x = 1 ± 2
   Answer: C. e^2 = 1 − (1/3)/(1/2) = 1/3; directrix X = ±a/e = ±sqrt(3/2) = ±sqrt6/2, so x = 1 ± sqrt6/2. (Key point: Eccentricity and directrices)

## Close variants set in other exams

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