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

# Sketch conic r = 10/(3+2cosθ)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2072 Kartik; 2072 Chaitra; 2067 Asar (asked 3 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-10-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 = 10/(3 + 2cosθ).

## Short answer

Rewrite as r=(10/3)/(1+(2/3)cosθ). It is an ellipse with eccentricity 2/3, directrix x=5, vertices at (2,0) and (−10,0), centre (−4,0).

## Answer

Write the equation in the standard polar form r = l/(1 + e cosθ), with the pole at a focus.

r = 10/(3 + 2cosθ) = (10/3) / (1 + (2/3)cosθ)

So e = 2/3 < 1, and the conic is an **ellipse**, with semi-latus rectum l = 10/3.

#### Key facts
- ed = l, so d = (10/3)/(2/3) = 5: directrix x = 5 (to the right of the pole).
- The major axis lies along the x-axis (cosθ term).
- θ = 0: r = 10/5 = 2, vertex (2, 0).
- θ = π: r = 10/1 = 10, vertex (−10, 0).
- θ = π/2: r = 10/3, so the ends of the latus rectum are (0, ±10/3).

#### Size of the ellipse
- Major axis 2a = 2 + 10 = 12, so a = 6.
- Centre = mid-point of the vertices = (−4, 0). Focal distance ae = 6 · 2/3 = 4, which matches: the pole (focus) is 4 units from the centre.
- b² = a²(1 − e²) = 36 · 5/9 = 20, so b = 2√5 ≈ 4.47.
- Other focus: (−8, 0).

```text
              y
              |        x = 5
          ____|____    |
        .'    |    '.  |
  -----(--F2--C--F1)--------- x
     -10 -8  -4  0 2
        '.____|____.'
              |
```

Sketch: a horizontal ellipse with vertices (−10, 0) and (2, 0), centre C(−4, 0), top and bottom at (−4, ±2√5), one focus F1 at the origin, the other F2 at (−8, 0), and the directrix x = 5 as a vertical line to the right of the curve.

## Check yourself

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

1. How do we put r = 10/(3 + 2cosθ) in the form r = l/(1 + e cosθ)?
   A. Multiply by 3, so e = 6
   B. Divide numerator and denominator by 3, so e = 2/3
   C. Divide by 2, so e = 3/2
   D. Put θ = 0
   Answer: B. r = (10/3)/(1 + (2/3)cosθ), so e = 2/3 and l = 10/3. (Key point: Standard polar form)

2. Why is the curve an ellipse?
   A. e = 2/3 < 1
   B. e = 2/3 so it is a parabola
   C. l = 10/3 > 1
   D. The denominator has cosθ
   Answer: A. e < 1 gives an ellipse. (Key point: Type of conic)

3. What is the directrix?
   A. x = 10/3
   B. x = −5
   C. y = 5
   D. x = 5
   Answer: D. ed = l gives d = (10/3)/(2/3) = 5, and the +cosθ sign puts the directrix at x = +5. (Key point: Directrix)

4. Vertices are at θ = 0 and θ = π. What is the major axis?
   A. 8
   B. 10
   C. 12, since r = 2 and r = 10 give 2a = 12
   D. 2
   Answer: C. r(0) = 2 and r(π) = 10, so 2a = 12 and a = 6. (Key point: Vertices and size)

## 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×
- [Sketch conic r = 10/(2−3sinθ)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-10-over-2-minus-3sin) — Mathematics I, chapter 3
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