If `theta` is the measure of acute angle between the pair of line repseented by `ax^(2) + 2hxy + by^(2) = 0` , then prove that
`tan theta = |(2sqrt(h^(2) - ab))/(a+b)|,a + b ne 0 `
Hence find the acture angle between the lines `x^(2) - 4xy + y^(2) = 0`

Explanation

Let `m_(1)` and `m_(2)` be the slopes of the lines represented by the equation `ax^(2) + 2hxy + by^(2) = 0." "…(1)`
Then their separate equations re `y = m_(1)x and y = m_(2)x.`
`therefore` their combined equation is `(m_(1)x -y) (m_(2)x -y) = 0`
i.e., `m_(1)m_(2)x^(2) = (m_(1) + m_(2)) xy + y^(2) = 0 " "...(2)`
Since (1) and (2) represent the same two lines, comparing the coefficients, we have,
`(m_(1)m_(2))/(2) =(-(m_(1)+m_(2)))/(2h) = (1)/(b) therefore m_(1)+m_(2) =-(2)/(b) and m_(1)m_(2) = (a)/(b)`
`therefore (m_(1)-m_(2))^(2) = (m_(1)+m_(2))^(2)-4m_(1)m_(2) = (4h^(2))/(b^(2)) -(4a)/(b) = (4(h^(2)-ab))/(b^(2))`
`therefore |m_(1)-m_(2)| = |(2sqrt(h^(2)-ab))/(b)|`
IF `theta` is the acute angle between the lines, then
`tan theta = |(m_(1)-m_(2))/(1+m_(1)m_(2))|, if m_(1)m_(2) ne -1`
`= |(2sqrt(h^(2)-ab//b))/(1+(a//b))|, if (a)/(b) ne -1`
` = |(2sqrt(h^(2)-ab))/(a+b)|, if a +b ne 0`.
(i) If the lines are perpendicular to each other.
Then `m_(1)m_(2) =-1 therefore (a)/(b) =-1 thereforea = -b therefore a + b = 0`
i.e., (coeff. of `x^(2)`) + (coeff. of `y^(2)`) = 0.
This is the condition for the lines to be perpendicular to each other.
(ii) If the lines are prallel (coincident), then the angle `theta` between them is 0.
`therefore tan theta = 0 therefore (2sqrt(h^(2)-ab))/(a+b) =0 therefore h^(2) -ab = 0`
This is condition for the lines to be parallel (coincident).