---
title: "Sketch conic r = 4secθ/(2secθ−1)"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:polar-conic-4sec-over-2sec-minus-1"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/polar-conic-4sec-over-2sec-minus-1
site: BCT Engineering
---

# Sketch conic r = 4secθ/(2secθ−1)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2069 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-4sec-over-2sec-minus-1
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the graph of the equation r = 4secθ/(2secθ − 1).

## Short answer

Multiply by cosθ: r = 2/(1 − ½ cosθ), e=½, ellipse. Pole is focus, directrix x=−4. Vertices (4,0) and (−4/3,0); centre (4/3,0).

## Answer

Multiply numerator and denominator by cosθ:

r = 4 secθ/(2 secθ − 1) = 4/(2 − cosθ) = 2/(1 − (1/2)cosθ)

Compare with r = l/(1 − e cosθ): e = 1/2 < 1, so the conic is an **ellipse**; l = 2.

#### Key facts
- ed = 2, so d = 4. With (1 − e cosθ) the directrix is on the left: x = −4.
- The major axis is along the x-axis.
- θ = 0: r = 4/(2 − 1) = 4, vertex (4, 0).
- θ = π: r = 4/3, vertex (−4/3, 0).
- θ = π/2: r = 2, so the latus rectum ends are (0, ±2).
- Major axis 2a = 4 + 4/3 = 16/3, so a = 8/3.
- Centre = (4/3, 0); ae = 4/3 ✓ (the pole is a focus); other focus (8/3, 0).
- b² = a²(1 − e²) = (64/9)(3/4) = 16/3, so b = 4/√3 ≈ 2.31.

```text
           y
    x=-4   |  (0,2)
     |   .-'-.
     |  / F   \
 ----|-(-+--C--)--x
     | (-4/3  4/3  4)
     |  \     /
        '-.-'
```

Sketch: an ellipse from x = −4/3 to x = 4 on the x-axis, centre (4/3, 0), passing through (0, ±2), with the focus at the origin and the directrix x = −4 on the left.

## Check yourself

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

1. How do we remove sec?
   A. Multiply by sinθ
   B. Divide by cosθ
   C. Multiply top and bottom by cosθ, giving 4/(2 − cosθ)
   D. Replace sec by 2
   Answer: C. secθ cosθ = 1. (Key point: Simplification)

2. What are e and the conic?
   A. e = 2, hyperbola
   B. e = 1/2, ellipse
   C. e = 1, parabola
   D. e = 1/2, hyperbola
   Answer: B. r = 2/(1 − (1/2)cosθ). (Key point: Type)

3. Where is the directrix?
   A. x = −4
   B. x = 4
   C. y = 4
   D. x = −2
   Answer: A. d = l/e = 4; the minus sign puts it on the left. (Key point: Directrix)

4. Vertices and centre?
   A. (2,0) and (−2,0); centre origin
   B. (4,0) and (4/3,0); centre (8/3,0)
   C. (0,2) and (0,−2); centre origin
   D. (4,0) and (−4/3,0); centre (4/3,0)
   Answer: D. r(0) = 4, r(π) = 4/3; 2a = 16/3, centre at the mid-point 4/3. (Key point: Vertices and centre)

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