Axes along bisectors of 7x²+4xy+4y²=0

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.