---
title: "Rotate axes by tan⁻¹2 in 4xy−3x²−a²=0"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:rotate-axes-4xy-3x2-tan-inverse-2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/rotate-axes-4xy-3x2-tan-inverse-2
site: BCT Engineering
---

# Rotate axes by tan⁻¹2 in 4xy−3x²−a²=0

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2072 Chaitra; 2070 Asar (asked 2 times)
- Video: [The equation 4 x y-3 x^2=a^2 become when the axes are turned through an angle tan ^-1 2 is (a) x^...](https://www.youtube.com/watch?v=YLrEbH62XHQ) — PW Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/rotate-axes-4xy-3x2-tan-inverse-2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

If the axes are turned through an angle tan⁻¹2, what does the equation 4xy − 3x² − a² = 0 become?

## Short answer

Using θ = tan⁻¹2, the equation 4xy − 3x² − a² = 0 transforms to X² − 4Y² = a², a hyperbola with transverse semi-axis a and conjugate semi-axis a/2.

## Answer

The axes are turned through θ = tan⁻¹2, so tan θ = 2, cos θ = 1/√5, sin θ = 2/√5.

x = X cosθ − Y sinθ = (X − 2Y)/√5
y = X sinθ + Y cosθ = (2X + Y)/√5

#### Substitution
4xy − 3x² = x(4y − 3x)

4y − 3x = [4(2X + Y) − 3(X − 2Y)]/√5 = (5X + 10Y)/√5

So

x(4y − 3x) = (X − 2Y)(5X + 10Y)/5 = (X − 2Y)(X + 2Y) = X² − 4Y²

The equation 4xy − 3x² − a² = 0 therefore becomes

**X² − 4Y² = a²**, that is X²/a² − Y²/(a/2)² = 1.

The xy term is removed and the curve is a hyperbola with transverse semi-axis a and conjugate semi-axis a/2.

## Check yourself

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

1. If θ = tan⁻¹2, what are cosθ and sinθ?
   A. 2/sqrt5 and 1/sqrt5
   B. 1/2 and 2
   C. 1/sqrt5 and 2/sqrt5
   D. 1/sqrt3 and 2/sqrt3
   Answer: C. tanθ = 2 gives a right triangle with sides 1, 2, sqrt5. (Key point: Trig values)

2. What are x and y in terms of X, Y?
   A. x = (X + 2Y)/sqrt5, y = (2X − Y)/sqrt5
   B. x = (X − 2Y)/sqrt5, y = (2X + Y)/sqrt5
   C. x = (2X − Y)/sqrt5, y = (X + 2Y)/sqrt5
   D. x = X − 2Y, y = 2X + Y
   Answer: B. x = X cosθ − Y sinθ, y = X sinθ + Y cosθ. (Key point: Substitution)

3. Why is 4xy − 3x^2 convenient to write as x(4y − 3x)?
   A. It gives (X − 2Y)(X + 2Y) = X^2 − 4Y^2 after substitution
   B. It removes a^2
   C. It makes x = 0
   D. It gives a parabola
   Answer: A. x(4y − 3x) = (X − 2Y)(5X + 10Y)/5 = X^2 − 4Y^2. (Key point: Factorising)

4. What does the equation become?
   A. X^2 + 4Y^2 = a^2
   B. 4X^2 − Y^2 = a^2
   C. X^2 − 4Y^2 = 0
   D. X^2 − 4Y^2 = a^2 (a hyperbola)
   Answer: D. The xy term vanishes and X^2/a^2 − Y^2/(a/2)^2 = 1. (Key point: Result)

## Close variants set in other exams

- [Rotate axes 45° in 3x²+3y²+2xy=2](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Arotate-axes-3x2-3y2-2xy-45) — Mathematics I, chapter 3, asked 5×
- [Rotate axes 30° in x²+2√3xy−y²=2a²](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Arotate-axes-x2-2sqrt3xy-y2-30) — Mathematics I, chapter 3, asked 5×

## Related in Mathematics I

- [Equation and length of axes of x²−4xy−2y²+10x+4y=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-x2-4xy-2y2-equation-axes) — Mathematics I, chapter 3
- [Axes along bisectors of 7x²+4xy+4y²=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Atransform-pair-of-lines-7x2-4xy-4y2) — Mathematics I, chapter 3
- [Centre, axes of 9x²+4xy+6y²−22x−16y+9=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-9x2-4xy-6y2-centre-axes) — Mathematics I, chapter 3, asked 7×
- [Axes of conic 14x²−4xy+11y²−44x−58y+71=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-14x2-4xy-11y2-axes) — Mathematics I, chapter 3, asked 2×
- [Product of semi-axes of x²−4xy+5y²=2](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-x2-4xy-5y2-product-semi-axes) — Mathematics I, chapter 3
- [Centre, axes, directrices of 3x²+8xy−3y²−40x−20y+50=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aconic-3x2-8xy-3y2-directrices) — Mathematics I, chapter 3, asked 7×

## Related in other subjects

- [Planes through line parallel to axes](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aplanes-through-line-parallel-axes) — Mathematics, chapter 3, asked 2×
- [Difference equation x(k+2)+3x(k+1)+2x(k)=0](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adifference-equation-x0-0-x1-1) — Digital Control System, chapter 2, asked 5×