Question

Prove that the homogeneous equation of degree two in \( x \) and \( y \), \( ax^2 + 2hxy + by^2 = 0 \), represents a pair of lines passing through the origin if \( h^2 - ab \geq 0 \). Hence, show that the equation \( x^2 + y^2 = 0 \) does not represent a pair of lines.

Hint:

To determine if a homogeneous quadratic equation represents a pair of lines, check the discriminant \( h^2 - ab \). If it's non-negative, the equation represents a pair of lines.

Answer 1

The general form of a homogeneous quadratic equation in \( x \) and \( y \) is: \[ ax^2 + 2hxy + by^2 = 0 \] This equation represents a pair of lines passing through the origin if and only if the discriminant \( \Delta = h^2 - ab \geq 0 \). To see why, we attempt to factorize the quadratic equation.

Step 1: Factorize the quadratic equation.
The quadratic equation can be factored as: \[ (ax + hy)(bx + hy) = 0 \] Expanding: \[ abx^2 + (ah + bh)xy + bhy^2 = 0 \] For this to match the given equation \( ax^2 + 2hxy + by^2 = 0 \), we must have \( 2h = ah + bh \), which simplifies to: \[ h^2 - ab \geq 0 \] Thus, the condition \( h^2 - ab \geq 0 \) ensures that the quadratic equation represents a pair of lines.

Step 2: Prove that \( x^2 + y^2 = 0 \) does not represent a pair of lines.
The equation \( x^2 + y^2 = 0 \) is a degenerate case where both \( x^2 \) and \( y^2 \) are non-negative, and their sum can only be zero if \( x = 0 \) and \( y = 0 \). This does not represent two distinct lines, so it does not satisfy the condition for a pair of lines.

Final Answer: The equation \( ax^2 + 2hxy + by^2 = 0 \) represents a pair of lines passing through the origin if and only if \( h^2 - ab \geq 0 \). The equation \( x^2 + y^2 = 0 \) does not represent a pair of lines.