---
title: "Sketch conic r = 12/(3+2cosθ)"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:polar-conic-12-over-3-plus-2cos"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/polar-conic-12-over-3-plus-2cos
site: BCT Engineering
---

# Sketch conic r = 12/(3+2cosθ)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2074 Chaitra, 5 marks; 2070 Asar (asked 2 times)
- Video: [Describe and Sketch the Conic | Plane Analytical Geometry | Engineering Mathematic 1st year 1st part](https://www.youtube.com/watch?v=G8Xn6zSks2o) — Padhne Keta
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/polar-conic-12-over-3-plus-2cos
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the graph of the conic r = 12/(3 + 2cosθ).

## Short answer

The conic simplifies to r=4/(1+(2/3)cosθ), eccentricity 2/3, so it is an ellipse with pole at a focus. Major axis length 14.4, centre (−4.8,0), foci at (0,0) and (−9.6,0), directrix x=6.

## Answer

Divide the numerator and denominator by 3 to match r = l/(1 + e cosθ):

r = 12/(3 + 2cosθ) = 4 / (1 + (2/3)cosθ)

So e = 2/3 < 1: the conic is an **ellipse**, with the pole at one focus and semi-latus rectum l = 4.

#### Key facts
- ed = 4, so d = 6: directrix x = 6.
- θ = 0: r = 12/5 = 2.4, vertex (2.4, 0).
- θ = π: r = 12, vertex (−12, 0).
- θ = π/2: r = 4, so the latus rectum ends are (0, ±4).
- Major axis 2a = 2.4 + 12 = 14.4, so a = 36/5 = 7.2.
- Centre = (−24/5, 0) = (−4.8, 0); ae = 4.8, so the other focus is (−9.6, 0).
- b² = a²(1 − e²) = (1296/25)(5/9) = 144/5, so b = 12/√5 ≈ 5.37.

```text
                              y
                              |          x = 6
              .  -  -  -  -  -|-  .        |
          .'                  | (0,4) '.   |
---------(-----F2-------C-----F1-------)---|------ x
       -12   -9.6     -4.8    0       2.4  6
          '.                  | (0,-4).'   |
              '  -  -  -  -  -|-  '        |
                              |
```

The curve is an ellipse stretched along the x-axis from x = −12 to x = 2.4, with centre C(−4.8, 0), foci F1 at the origin (the pole) and F2(−9.6, 0), passing through (0, ±4), and with the directrix x = 6 on the right.

## Check yourself

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

1. What are e and l for r = 12/(3 + 2cosθ)?
   A. e = 2, l = 12
   B. e = 2/3, l = 4
   C. e = 3/2, l = 4
   D. e = 2/3, l = 12
   Answer: B. Dividing by 3 gives r = 4/(1 + (2/3)cosθ). (Key point: Standard form)

2. What is the directrix?
   A. x = 6
   B. x = 4
   C. x = 8/3
   D. x = −6
   Answer: A. d = l/e = 4/(2/3) = 6, to the right of the pole. (Key point: Directrix)

3. Where are the vertices?
   A. (4, 0) and (−4, 0)
   B. (12, 0) and (2.4, 0)
   C. (0, 4) and (0, −4)
   D. (2.4, 0) and (−12, 0)
   Answer: D. r(0) = 12/5 = 2.4 and r(π) = 12/1 = 12 (in the direction π). (Key point: Vertices)

4. What are the major axis and centre?
   A. 2a = 12; centre (−6, 0)
   B. 2a = 9.6; centre (−2.4, 0)
   C. 2a = 14.4; centre (−4.8, 0)
   D. 2a = 14.4; centre (0, 0)
   Answer: C. 2a = 2.4 + 12 = 14.4, centre is the mid-point of vertices (−4.8, 0), with focus at the origin. (Key point: Size)

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