Question

Show that the acute angle θ between the lines represented by ax² + 2hxy + by² = 0 is given by, tan θ = `|(2sqrt(h^2 - ab))/(a + b)|`.

Answer 1

Let m₁ and m₂ be the slopes of the lines represented by the equation ax² + 2hxy + by² = 0.   ...(1)

Then their separate equations are y = m₁x and y = m₂x

∴ Their combined equation is (m1x – y)(m2x – y) = 0

i.e. m₁m₂x² – (m1 + m2) xy + y2= 0   ...(2)

Since (1) and (2) represent the same two lines, comparing the coefficients, we get

`(m_1m_2)/a = (-(m_1 + m_2))/(2h) = 1/b`

∴ `m_1 + m_2 = - (2h)/b` and `m_1m_2 = 1/b`

∴ (m₁ –  m₂)² = (m₁ + m₂)² – 4m₁m₂

= `(4h^2)/(b^2) - (4a)/b`

= `(4(h^2 - ab))/(b^2)`

∴ `|m_1 - m_2| = |(2sqrt(h^2 - ab))/b|`

If θ is the acute angle between the lines, then

`tan θ = |(m_1 - m_2)/(1 + m_1m_2)|`, if m₁m₂ ≠ –1

= `|((2sqrt(h^2 - ab))//b)/(1 + (a//b))|`, if `a/b ≠ -1`

∴ `tan θ = |(2sqrt(h^2 - ab))/(a + b)|`, if a + b ≠ 0.