---
title: "Sketch conic r = 10cosecθ/(2cosecθ+3)"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:polar-conic-10cosec-over-2cosec-plus-3"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/polar-conic-10cosec-over-2cosec-plus-3
site: BCT Engineering
---

# Sketch conic r = 10cosecθ/(2cosecθ+3)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2073 Chaitra
- 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-10cosec-over-2cosec-plus-3
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the conic r = 10cosecθ/(2cosecθ + 3).

## Short answer

Multiply by sinθ: r = 5/(1 + 1.5 sinθ), e=1.5, hyperbola. Pole is focus, directrix y=10/3. Vertices (0,2) and (0,10); centre (0,6).

## Answer

Multiply the numerator and denominator by sinθ (cosecθ · sinθ = 1):

r = 10 cosecθ/(2 cosecθ + 3) = 10/(2 + 3 sinθ) = 5/(1 + (3/2) sinθ)

Compare with r = l/(1 + e sinθ): e = 3/2 > 1, so the conic is a **hyperbola**; l = 5.

#### Key facts
- ed = 5, so d = 10/3: directrix y = 10/3.
- The transverse axis is along the y-axis.
- θ = π/2: r = 10/5 = 2, vertex (0, 2).
- θ = 3π/2: r = 10/(2 − 3) = −10, giving the vertex (0, 10) (a negative r reverses the direction).
- Centre = (0, 6); a = 4; ae = 6, so the foci are (0, 0) and (0, 12).
- b² = a²(e² − 1) = 16 · 5/4 = 20, so b = 2√5.
- Asymptotes: 1 + (3/2)sinθ = 0, sinθ = −2/3, θ ≈ 221.8° and 318.2° (r → ∞). Through the centre their slopes are ±a/b = ±2/√5.
- θ = 0 or π: r = 5, so the latus rectum ends are (±5, 0).

```text
                       y
         \             |             /
          \            * F2 (0,12)  /
           '.          |          .'
             '-.______ V' ______.-'      upper branch, vertex V'(0,10)
                       |
                       + C (0,6)         centre
                       |
   - - - - - - - - - - | - - - - - - - - directrix y = 10/3
              .-'''''' V ''''''-.        lower branch, vertex V(0,2)
           .'          |          '.
   -------/------------F------------\------- x   F = pole (0,0)
       (-5,0)          |          (5,0)
         /             |             \
```

Sketch: a hyperbola opening up and down, with vertices V(0, 2) and V'(0, 10) and centre C(0, 6). The lower branch (vertex V) curves around the focus at the pole and crosses the x-axis at (±5, 0); the upper branch (vertex V') curves around the focus F2(0, 12). The branches are not mirror images about the x-axis; they are symmetric about the centre (0, 6). The directrix y = 10/3 lies between the vertex V and the centre C.

## Check yourself

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

1. How is the cosec removed?
   A. Multiply by cosθ
   B. Replace cosec by 1
   C. Multiply top and bottom by sinθ giving r = 10/(2 + 3 sinθ)
   D. Differentiate
   Answer: C. cosecθ · sinθ = 1. (Key point: Simplification)

2. What is e and the conic?
   A. e = 2/3, ellipse
   B. e = 3/2, hyperbola
   C. e = 3, hyperbola
   D. e = 1, parabola
   Answer: B. r = 5/(1 + (3/2)sinθ). (Key point: Type)

3. What are the vertices?
   A. (0, 2) and (0, 10)
   B. (0, 2) and (0, −10)
   C. (2, 0) and (10, 0)
   D. (0, −2) and (0, −10)
   Answer: A. r(π/2) = 2; r(3π/2) = −10, giving (0, 10). (Key point: Vertices)

4. What are centre, a and foci?
   A. Centre (0, 4), a = 6
   B. Centre (0, −6), a = 4
   C. Centre (0, 6), a = 8
   D. Centre (0, 6), a = 4, foci (0,0) and (0,12)
   Answer: D. Vertices 2 and 10 give a = 4, centre 6, ae = 6. (Key point: Centre and a)

## Related in Mathematics I

- [Sketch polar conic r = 10cosecθ/(...)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-10cosec-denominator-cut) — Mathematics I, chapter 3