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

# Sketch polar conic r = 10cosecθ/(...)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2070 Chaitra
- Video: [Graphing Conic Sections Using Polar Equations - Part 1](https://www.youtube.com/watch?v=zS7ZjWxu8CY) — Mathispower4u
- 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-denominator-cut
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the graph of the polar equation of conic r = 10cosecθ/[denominator cut off in the scan].

## Short answer

Multiply by sinθ to get r=(10/m)/(1+(n/m)sinθ). Eccentricity e=n/m. If n<m ellipse, n=m parabola, n>m hyperbola. Focus at pole, directrix y=10/n.

## Answer

**Note:** the denominator is cut off in the scan, so the exact curve cannot be fixed. The paper's form is r = 10 cosecθ/(m cosecθ + n); the method below works for any constants m, n > 0, and the result is read off from n/m.

#### Step 1: convert to standard polar form
Multiply the numerator and denominator by sinθ (cosecθ · sinθ = 1):

r = 10/(m + n sinθ) = (10/m) / (1 + (n/m) sinθ)

Compare with r = l/(1 + e sinθ):

- eccentricity **e = n/m**
- semi-latus rectum l = 10/m
- directrix: ed = l, so d = 10/n, and the directrix is y = 10/n (above the pole)
- the pole is a focus and the axis of the conic is the y-axis

#### Step 2: identify the conic
| Condition | e | Conic |
|---|---|---|
| n < m | < 1 | Ellipse |
| n = m | = 1 | Parabola |
| n > m | > 1 | Hyperbola |

#### Step 3: vertices
- θ = π/2: r = 10/(m + n), vertex (0, 10/(m + n)).
- θ = 3π/2: r = 10/(m − n), vertex (0, −10/(m − n)).
- θ = 0 and π: r = 10/m, the ends of the latus rectum (±10/m, 0).

#### Step 4: sketch
- For n < m: an ellipse with major axis along the y-axis, a = 10m/(m² − n²), centre (0, −10n/(m² − n²)), b² = a²(1 − e²), one focus at the origin.
- For n = m: a parabola opening downward, with focus at the origin and directrix y = 10/n.
- For n > m: a hyperbola with transverse axis on the y-axis.

**Worked example (m = 3, n = 2):** r = 10/(3 + 2 sinθ), e = 2/3, ellipse; vertices (0, 2) and (0, −10), so a = 6, centre (0, −4), b² = 20; directrix y = 5.

```text
          y
          |   y = 5
  - - - - | - - - - - -
          | V(0,2)
       .--+--.
      /   F   \
  ---(----+----)---- x
      \   C   /
       '--.--'
          V'(0,-10)
```

## Check yourself

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

1. How is r = 10cosecθ/(m cosecθ + n) put in standard form?
   A. Multiply by cosθ
   B. Put m = n
   C. Multiply by sinθ: r = (10/m)/(1 + (n/m) sinθ)
   D. Set θ = π/2
   Answer: C. This gives e = n/m and l = 10/m. (Key point: Conversion)

2. Which condition gives a hyperbola?
   A. n < m
   B. n > m (e > 1)
   C. n = m
   D. m = 0
   Answer: B. e = n/m > 1 for a hyperbola; e < 1 ellipse; e = 1 parabola. (Key point: Classification)

3. Where is the directrix?
   A. y = 10/n
   B. x = 10/n
   C. y = 10/m
   D. y = −10/n
   Answer: A. ed = l gives d = 10/n and the +sinθ puts it above the pole. (Key point: Directrix)

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