Question

Let S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0 where h² > ab represent a pair of lines both passing through origin. Prove that g = f = c = 0.

Answer 1

Since S = 0 passes through (0, 0) we have c = 0. Let P(x₁, y₁) and Q(x₂, y₂) be points on one of the lines other than the origin so that area of  ΔOPQ ≠ 0. This implies that

P(x₁, y₁) lies on one line which passes through origin. This implies that (-x₁, y₁) also lies on the line. Therefore

 ax²₁ + 2hx₁y₁ + by₁² + 2gx₁ + 2fy₁ = 0

and   ax²₁ + 2hx₁y₁ + by₁² + 2gx₁ - 2fy₁ = 0

 imply that

 gx₁ + fy₁ = 0   ....(2)

 Similarly

 gx₂ + fy₂ = 0   ....(3)

 Therefore, g = f = 0 because x₁y₂ - x₂y₁ ≠ 0 or the matrix

is a non-singular matrix. Equations (2) and (3) have zero solution only so that g = 0 and f = 0.