---
title: "Centre, axes, directrices of 3x²+8xy−3y²−40x−20y+50=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:conic-3x2-8xy-3y2-directrices"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-3x2-8xy-3y2-directrices
site: BCT Engineering
---

# Centre, axes, directrices of 3x²+8xy−3y²−40x−20y+50=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2073 Shrawan; 2068 Shrawan (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/conic-3x2-8xy-3y2-directrices
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the centre, length of axes, eccentricity and directrices of the conic 3x² + 8xy − 3y² − 40x − 20y + 50 = 0.

## Short answer

Centre (4,2); after rotation, standard form u²/10−v²/10=1. Axes lengths 2√10 each; eccentricity √2; directrices 2x+y−5=0 and 2x+y−15=0.

## Answer

Here a = 3, h = 4, b = −3, so h² − ab = 25 > 0 and a + b = 0: it is a **rectangular hyperbola**.

#### Centre
3x + 4y − 20 = 0 and 4x − 3y − 10 = 0. Solving: x = 4, y = 2. **Centre: (4, 2).**

#### Shift origin to the centre
Constant term: g·x₀ + f·y₀ + c = (−20)(4) + (−10)(2) + 50 = −50, so

3X² + 8XY − 3Y² = 50

#### Rotate
tan 2θ = 2h/(a − b) = 8/6 = 4/3, so cos 2θ = 3/5, tan θ = 1/2: cos θ = 2/√5, sin θ = 1/√5. The new coefficients are the roots of λ² − (a+b)λ + (ab − h²) = 0, i.e. λ² − 25 = 0, λ = ±5. Thus

5u² − 5v² = 50, or **u²/10 − v²/10 = 1**

where u is measured along the direction (2, 1)/√5.

#### Results
- a² = b² = 10, so the semi-axes are √10 each: **length of axes = 2√10 (both)**.
- **Eccentricity** e = √(1 + b²/a²) = **√2**.
- Directrices in the new frame: u = ±a/e = ±√5. Since u = [2(x−4) + (y−2)]/√5 = (2x + y − 10)/√5,

2x + y − 10 = ±5

**Directrices: 2x + y − 5 = 0 and 2x + y − 15 = 0.**

- Axes: transverse x − 2y = 0; conjugate 2x + y − 10 = 0 (both through (4, 2)).
- Foci: u = ±ae = ±2√5, i.e. (4, 2) ± (4, 2) = (8, 4) and (0, 0).

## Check yourself

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

1. For a=3,h=4,b=−3 why is it a rectangular hyperbola?
   A. h^2 − ab < 0 and a = b
   B. h^2 − ab = 25 > 0 and a + b = 0
   C. a + b = 6
   D. h = 0
   Answer: B. A hyperbola has h^2 − ab > 0; a + b = 0 makes the asymptotes perpendicular. (Key point: Type of conic)

2. What is the centre?
   A. (4, 2)
   B. (2, 4)
   C. (−4, −2)
   D. (0, 0)
   Answer: A. 3x + 4y − 20 = 0 and 4x − 3y − 10 = 0 give x = 4, y = 2. (Key point: Centre)

3. How do we get u^2/10 − v^2/10 = 1?
   A. Roots of λ^2 + 25 = 0
   B. Use λ = 3 and −3
   C. Divide by h
   D. The new coefficients are roots of λ^2 − 25 = 0, i.e. ±5, giving 5u^2 − 5v^2 = 50
   Answer: D. Roots of λ^2 − (a+b)λ + (ab − h^2) = 0 are ±5. (Key point: Rotated form)

4. What are e and the directrices?
   A. e = 2; 2x + y = 10
   B. e = sqrt2; x − 2y = 0
   C. e = sqrt2; 2x + y − 5 = 0 and 2x + y − 15 = 0
   D. e = 1; 2x + y − 10 = 0
   Answer: C. a = b gives e = sqrt2; directrices u = ±a/e = ±sqrt5, i.e. 2x + y − 10 = ±5. (Key point: Eccentricity and directrices)

## Close variants set in other exams

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