Question

If the equation ax^2 + 2hxy + by^2 = 0 represents a pair of straight lines, the prove that the angle θ between the lines is given by cos θ = |a + b|/√(a + b)^2 + 4h^2​

Answer 1

Answer:

Compare given equation with ax2+2hxy+by2+2gx+2fy+c=0 we have,

a=6b=4h=25

θ=tan−1(m) (given)

=tan−1{a+b2h2−ab}=tan−1⎩⎪⎨⎪⎧6−42425+24⎭⎪⎬⎪⎫

=tan−1{211}   ∴m=211

⇒(2m=11)               .............(*)

Again a2+b2−ab−a−b+1≤0

⇒21(2a2+2b