What are x and y in terms of X, Y?

x = (X − 2Y)/sqrt5, y = (2X + Y)/sqrt5

x = (X + 2Y)/sqrt5, y = (2X − Y)/sqrt5

x = (2X − Y)/sqrt5, y = (X + 2Y)/sqrt5

Why is 4xy − 3x^2 convenient to write as x(4y − 3x)?

It gives (X − 2Y)(X + 2Y) = X^2 − 4Y^2 after substitution

X^2 − 4Y^2 = a^2 (a hyperbola)

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.

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

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

4xy − 3x² = x(4y − 3x)

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

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.

✓ Answer checked by Claude Opus 5.5 (high effort)