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

# Sketch conic r = 10/(2−3sinθ)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2071 Chaitra
- 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-10-over-2-minus-3sin
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the graph of the equation r = 10/(2 − 3sinθ).

## Short answer

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

## Answer

Divide by 2:

r = 10/(2 − 3 sinθ) = 5/(1 − (3/2) sinθ)

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

#### Key facts
- ed = 5, so d = 10/3. With (1 − e sinθ) the directrix is below the pole: y = −10/3.
- The transverse axis is along the y-axis.
- θ = π/2: r = 10/(2 − 3) = −10, giving the vertex (0, −10).
- θ = 3π/2: r = 10/5 = 2, giving the vertex (0, −2).
- Centre = (0, −6); a = 4; ae = 6, so the foci are (0, 0) and (0, −12).
- b² = a²(e² − 1) = 20, b = 2√5.
- Asymptotes: sinθ = 2/3, θ ≈ 41.8° and 138.2°.
- θ = 0 or π: r = 5, so the latus rectum ends are (±5, 0).

```text
          y
          |
     ___  | F(0,0)  ___
        \_V(0,-2)_/    upper branch
          C(0,-6)
        _/V'(0,-10)\_  lower branch
     ___/ F2(0,-12) \___
```

Sketch: a hyperbola opening up and down (mirror image of r = 10/(2 + 3 sinθ) in the x-axis). Its vertices are (0, −2) and (0, −10), the centre is (0, −6), the foci are the origin and (0, −12), and the asymptotes make angles of about 41.8° and 138.2° with the x-axis through the centre.

## Check yourself

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

1. r = 10/(2 − 3sinθ) becomes?
   A. r = 5/(1 + (3/2)sinθ)
   B. r = 10/(1 − 3 sinθ)
   C. r = 5/(1 − (3/2)sinθ)
   D. r = 5/(1 − (2/3)sinθ)
   Answer: C. Divide by 2: e = 3/2, l = 5. (Key point: Standard form)

2. Where is the directrix?
   A. y = 10/3
   B. y = −10/3
   C. x = −10/3
   D. y = 5
   Answer: B. d = l/e = 10/3 and the minus sign puts the directrix below the pole. (Key point: Directrix)

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

4. How does it relate to r = 10/(2 + 3sinθ)?
   A. It is a rotation by 90°
   B. It is identical
   C. It is its reflection in the y-axis
   D. It is its mirror image in the x-axis
   Answer: D. Replacing sinθ by −sinθ reflects the curve in the x-axis; centre (0,−6). (Key point: Relation to previous curve)

## Related in Mathematics I

- [Sketch conic r = 10/(3+2cosθ)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-10-over-3-plus-2cos) — Mathematics I, chapter 3, asked 3×
- [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
- [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×
- [Sketch conic r = 4secθ/(2secθ−1)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-4sec-over-2sec-minus-1) — Mathematics I, chapter 3
- [Equation and length of axes of x²−4xy−2y²+10x+4y=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-x2-4xy-2y2-equation-axes) — Mathematics I, chapter 3

## Related in other subjects

- [Circular arch 40 m: sketch, equation, forces at 10 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-40m-60kn-4knm) — Theory of Structures I, chapter 6
- [Irrigation interval with AMC, RAM graph](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Ainterval-ram75-graph) — Irrigation and Drainage Engineering, chapter 2, asked 4×
- [Kruskal's MST on graph A to G](https://www.bctengineer.com/question.md?key=semester_4%2Fdata_structure_and_algorithm%3Akruskal-mst-graph-d) — Data Structure and Algorithm, chapter 10, asked 3×
- [Kruskal's MST on 9-vertex graph](https://www.bctengineer.com/question.md?key=semester_4%2Fdata_structure_and_algorithm%3Akruskal-mst-graph-f) — Data Structure and Algorithm, chapter 10, asked 3×