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

# Sketch conic r = 12/(2−6cosθ)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2073 Shrawan
- 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-2-minus-6cos
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the conic r = 12/(2 − 6cosθ).

## Short answer

r = 6/(1 − 3cosθ), e=3, hyperbola. Pole is focus, directrix x=−2. Vertices at (−3,0) and (−3/2,0).

## Answer

Divide by 2 to get the form r = l/(1 − e cosθ):

r = 12/(2 − 6cosθ) = 6/(1 − 3cosθ)

So e = 3 > 1: the conic is a **hyperbola**, with the pole at a focus and l = 6.

#### Key facts
- ed = 6, so d = 2. The sign (1 − e cosθ) puts the directrix to the left of the pole: x = −2.
- θ = 0: r = 6/(1 − 3) = −3, giving the vertex (−3, 0) (a negative r means the point lies in the opposite direction).
- θ = π: r = 6/4 = 3/2, giving the vertex (−3/2, 0).
- θ = π/2: r = 6, so the latus rectum ends are (0, ±6).
- Centre = (−9/4, 0); a = 3/4; ae = 9/4, so the foci are (0, 0) and (−9/2, 0).
- b² = a²(e² − 1) = (9/16)(8) = 9/2.
- Asymptote directions: 1 − 3cosθ = 0, cosθ = 1/3, θ = ±70.5° (r → ∞).

```text
                                                y
            \                       /           |
             \                     /            |
              \                   /             |
               )        :        (              |
 ----------F2--V-----C--:--------V'-------------F1------ x
               )        :        (              |
              /         :         \             |
             /          :          \            |
            /           :           \           |
                     x = -2
 F2(-9/2, 0)   V(-3, 0)   C(-9/4, 0)   V'(-3/2, 0)   F1 = pole (0, 0)
```

Sketch: a hyperbola opening left and right, with vertices V(−3, 0) and V'(−3/2, 0) and centre C(−9/4, 0). The right branch (vertex V') curves around the focus F1 at the pole and passes through (0, ±6); the left branch (vertex V) curves around F2(−9/2, 0). The asymptotes pass through C at about ±70.5° to the x-axis, and the directrix x = −2 lies between C and V'.

## Check yourself

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

1. What does r = 12/(2 − 6cosθ) become?
   A. r = 12/(1 − 6cosθ), e = 6
   B. r = 6/(1 − cosθ), e = 1
   C. r = 6/(1 − 3cosθ), e = 3
   D. r = 6/(1 + 3cosθ), e = 3
   Answer: C. Divide numerator and denominator by 2. (Key point: Standard form)

2. Why is it a hyperbola?
   A. e = 3 and l = 6
   B. e = 3 > 1
   C. The sign is negative
   D. Because d = 2
   Answer: B. e > 1 gives a hyperbola. (Key point: Type)

3. Where is the directrix?
   A. x = −2
   B. x = 2
   C. x = 6
   D. y = −2
   Answer: A. ed = 6 gives d = 2; the minus sign in 1 − e cosθ puts it to the left. (Key point: Directrix)

4. At θ = 0, r = −3. Where is that point?
   A. (3, 0)
   B. (0, −3)
   C. (0, 3)
   D. (−3, 0), since negative r goes the opposite direction
   Answer: D. A negative r is plotted in the opposite direction, giving (−3, 0). (Key point: Vertex with negative r)

## 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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