---
title: "Sketch conic r = 12/(6+2sinθ)"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:polar-conic-12-over-6-plus-2sin"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/polar-conic-12-over-6-plus-2sin
site: BCT Engineering
---

# Sketch conic r = 12/(6+2sinθ)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2074 Asoj
- Video: [Polar Conics Example 2 - Sketch r = 12/(6+2sinθ)](https://www.youtube.com/watch?v=DxpKguTdUes) — MasterWuMathematics
- 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-6-plus-2sin
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the conic r = 12/(6 + 2sinθ).

## Short answer

r = 2/(1 + ⅓ sinθ), e=⅓, ellipse. Pole is focus, directrix y=6. Vertices (0, 3/2) and (0, −3); centre (0, −¾).

## Answer

Divide by 6 to get the form r = l/(1 + e sinθ):

r = 12/(6 + 2sinθ) = 2/(1 + (1/3) sinθ)

So e = 1/3 < 1: an **ellipse**, with the pole at a focus and l = 2.

#### Key facts
- ed = 2, so d = 6: the directrix is the horizontal line y = 6.
- The major axis lies along the y-axis (sinθ term).
- θ = π/2: r = 12/8 = 3/2, vertex (0, 3/2).
- θ = 3π/2: r = 12/4 = 3, vertex (0, −3).
- θ = 0 or π: r = 2, so the latus rectum ends are (±2, 0).
- Major axis 2a = 3/2 + 3 = 9/2, so a = 9/4. Centre (0, −3/4); ae = 3/4 (the pole is 3/4 from the centre ✓).
- b² = a²(1 − e²) = (81/16)(8/9) = 9/2, so b = 3/√2 ≈ 2.12.
- Other focus: (0, −3/2).

```text
         y
         |      y = 6
  - - - - - - - - - - - -
         | (0,1.5)
       .-'-.
      /  F  \ (2,0)
 ----(---+---)----x
      \  C  /
       '-.-'
         | (0,-3)
```

The ellipse is long vertically, from y = −3 to y = 3/2, with a focus at the origin and the directrix y = 6 above it.

## Check yourself

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

1. r = 12/(6 + 2sinθ) in standard form?
   A. r = 2/(1 + 3 sinθ)
   B. r = 6/(1 + sinθ/6)
   C. r = 2/(1 + (1/3)sinθ)
   D. r = 12/(1 + sinθ)
   Answer: C. Divide by 6: e = 1/3, l = 2. (Key point: Standard form)

2. Why is the axis along the y-axis and the directrix horizontal?
   A. The term is sinθ, giving directrix x = 6
   B. The term is sinθ, giving directrix y = 6
   C. The term is cosθ
   D. Because e < 1
   Answer: B. sinθ terms give a horizontal directrix y = d = 6 and axis along y. (Key point: Orientation)

3. What are the vertices?
   A. (0, 3/2) and (0, −3)
   B. (0, 2) and (0, −2)
   C. (3/2, 0) and (−3, 0)
   D. (0, 6) and (0, −6)
   Answer: A. θ = π/2 gives r = 3/2; θ = 3π/2 gives r = 3 in the downward direction. (Key point: Vertices)

4. What are a and the centre?
   A. a = 9/2, centre (0, 0)
   B. a = 3, centre (0, −3)
   C. a = 9/4, centre (0, 3/4)
   D. a = 9/4, centre (0, −3/4)
   Answer: D. 2a = 3/2 + 3 = 9/2; centre is the mid-point (0, −3/4). (Key point: Centre and a)

## Related in Mathematics I

- [Sketch conic r = 12/(3+2cosθ)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-12-over-3-plus-2cos) — Mathematics I, chapter 3, asked 2×
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