---
title: "Sketch polar conic r = 12/(...) (2068 Chaitra)"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:polar-conic-12-denominator-cut-2068"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/polar-conic-12-denominator-cut-2068
site: BCT Engineering
---

# Sketch polar conic r = 12/(...) (2068 Chaitra)

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2068 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-12-denominator-cut-2068
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Describe and sketch the polar conic r = 12/[denominator cut off in the scan].

## Short answer

Rewrite as r = (12/p)/(1+(q/p)cosθ) to identify eccentricity e=q/p and semi-latus rectum l=12/p. The conic is an ellipse if q<p, parabola if q=p, hyperbola if q>p. Key points: vertices at r(0)=12/(p+q), r(π)=12/(p−q); latus rectum ends at (0,±12/p).

## Answer

**Note:** the denominator is cut off in the scan, so the exact curve cannot be fixed. The method below works for r = 12/(p + q cosθ) (or sinθ) with unknown constants p, q > 0; the answer is decided by the ratio q/p.

#### Step 1: standard form
r = 12/(p + q cosθ) = (12/p) / (1 + (q/p) cosθ)

This has the form r = l/(1 + e cosθ), so

- eccentricity **e = q/p**
- semi-latus rectum l = 12/p
- directrix at distance d = l/e = 12/q from the pole: x = 12/q
- the pole is at a focus and the axis is the x-axis

#### Step 2: classify
- q < p: ellipse (0 < e < 1)
- q = p: parabola (e = 1)
- q > p: hyperbola (e > 1)

#### Step 3: key points
- Put θ = 0: r₁ = 12/(p + q), the vertex (12/(p + q), 0).
- Put θ = π: r₂ = 12/(p − q), the vertex (−12/(p − q), 0).
- Put θ = π/2: r = 12/p, so the latus rectum has length 24/p, with ends (0, ±12/p).
- For an ellipse, 2a = r₁ + r₂ = 24p/(p² − q²); the centre is the mid-point of the vertices and b² = a²(1 − e²).

#### Step 4: draw
Draw the x- and y-axes, put the focus at the pole, plot the two vertices and the latus rectum ends, and join them by a smooth closed curve (ellipse), one open curve (parabola) or two branches (hyperbola). If the denominator contains sinθ instead of cosθ, the same working applies with the axis along the y-axis and the directrix y = 12/q.

**Example (p = 2, q = 2):** r = 12/(2 + 2cosθ) = 6/(1 + cosθ), a parabola with focus at the origin, vertex (3, 0), directrix x = 6, opening to the left.

## Check yourself

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

1. Using θ = π/2, what is the length of the latus rectum?
   A. 12/p
   B. 24/p
   C. 12/q
   D. 24/q
   Answer: B. r = 12/p at θ = π/2 and the full length is twice that. (Key point: Latus rectum)

2. For an ellipse, how is 2a found?
   A. 2a = r1 + r2 = 24p/(p^2 − q^2)
   B. 2a = r1 − r2
   C. 2a = 12/p
   D. 2a = 24/q
   Answer: A. The sum of the two vertex distances is the major axis. (Key point: Ellipse axis)

## Close variants set in other exams

- [Sketch polar conic r = 12/(...) (2071 Shrawan)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-12-denominator-cut-2071) — Mathematics I, chapter 3, asked 2×

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