---
title: "Axes along bisectors of 7x²+4xy+4y²=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:transform-pair-of-lines-7x2-4xy-4y2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/transform-pair-of-lines-7x2-4xy-4y2
site: BCT Engineering
---

# Axes along bisectors of 7x²+4xy+4y²=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2074 Asoj
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/transform-pair-of-lines-7x2-4xy-4y2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

What does the equation of lines 7x² + 4xy + 4y² = 0 become when the axes are the bisectors of the angles between them?

## Short answer

Find bisectors: (x²−y²)/(a−b)=xy/h → 2x²−3xy−2y²=0 → (2x+y)(x−2y)=0. Taking new X-axis along x−2y=0 (tanθ=1/2), substitute to get 8X²+3Y²=0.

## Answer

**Reading of the question:** the equation is taken as given, 7x² + 4xy + 4y² = 0. Here h² − ab = 4 − 28 < 0, so the two lines are imaginary (they meet only at the origin in the real plane), but their bisector directions are still real and the transformation works.

#### Bisectors of the angles between the lines
For ax² + 2hxy + by² = 0 the bisectors are (x² − y²)/(a − b) = xy/h. With a = 7, b = 4, h = 2:

(x² − y²)/3 = xy/2
2x² − 3xy − 2y² = 0
(2x + y)(x − 2y) = 0

So the bisectors are x − 2y = 0 and 2x + y = 0. They are perpendicular, as they must be.

#### New axes
Take the new X-axis along x − 2y = 0, direction (2, 1), so tanθ = 1/2, cosθ = 2/√5, sinθ = 1/√5:

x = (2X − Y)/√5, y = (X + 2Y)/√5

(The same angle satisfies tan 2θ = 2h/(a − b) = 4/3.)

#### Substitute
- x² = (4X² − 4XY + Y²)/5
- xy = (2X² + 3XY − 2Y²)/5
- y² = (X² + 4XY + 4Y²)/5

7x² + 4xy + 4y² = [28X² − 28XY + 7Y² + 8X² + 12XY − 8Y² + 4X² + 16XY + 16Y²]/5

= [40X² + 0·XY + 15Y²]/5 = 8X² + 3Y²

**The equation becomes 8X² + 3Y² = 0.**

(If the other bisector is taken as the X-axis, it becomes 3X² + 8Y² = 0.) Since both coefficients are positive, the lines are imaginary: only the origin satisfies the equation.

## Check yourself

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

1. Bisectors of ax^2 + 2hxy + by^2 = 0 are?
   A. (x^2 + y^2)/(a + b) = xy/h
   B. (x^2 − y^2)/h = xy/(a − b)
   C. x^2/a = y^2/b
   D. (x^2 − y^2)/(a − b) = xy/h
   Answer: D. Standard bisector formula. (Key point: Bisector equation)

2. What are the bisectors of 7x^2 + 4xy + 4y^2 = 0?
   A. x + 2y = 0 and 2x − y = 0
   B. x = 0 and y = 0
   C. x − 2y = 0 and 2x + y = 0
   D. x − y = 0 and x + y = 0
   Answer: C. 2x^2 − 3xy − 2y^2 = (2x + y)(x − 2y) = 0. (Key point: Bisectors here)

3. What does the equation become?
   A. 7X^2 + 4Y^2 = 0
   B. 8X^2 + 3Y^2 = 0
   C. 8X^2 − 3Y^2 = 0
   D. 15X^2 = 0
   Answer: B. Substitution gives (40X^2 + 15Y^2)/5. (Key point: Result)

4. What does 8X^2 + 3Y^2 = 0 show?
   A. The lines are imaginary; only the origin is real
   B. Two real perpendicular lines
   C. Coincident lines
   D. A circle
   Answer: A. Both coefficients positive means no real lines; h^2 − ab < 0. (Key point: Nature of lines)

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