Find the measure of the acute angle between the lines represented by `(a^(2) - 3b^(2)) x^(2) + 8abxy + (b^(2) - 3a^(2)) y^(2) = 0.`

Explanation

Comparing the equation
`(a^(2)-3b^(2))x^(2) + 8abxy + (b^(2) -3a
^(2)) y^(2) = -0`, with
`Ax^(2) + 2Hxy + By^(2) =0,` we have,
`A = a^(2) -3b^(2), H = 4ab, B = b^(2) -3a^(2).`
`therefore H^(2) -AB = 16a^(2)b^(2)-(a^(2)-3b^(2))(b^(2)-3a^(2))`
` = 16a^(2)b^(2) +(a^(2)-3b^(2))(3a^(2)-b^(2))`
`=16a^(2)b^(2)+3a^(4)-10a^(2)b^(2)+3b^(4)`
` = 3a^(4)+6a^(2)b^(2)+3b^(4)`
`=3(a^(4)+2a^(2)b^(2)+b^(4))`
` = 3(a^(2) + b^(2) )^(2)`
`therefore sqrt(H^(2)-AB) = sqrt(3)(a^(2)+b^(2))`
Also, `A + B = (a^(2)-3b^(2))+(b^(2)-3a^(2))=-2(a^(2)+b^(2))`
IF `theta` is the acute angle between the lines, then
`tan theta = |(2sqrt(H^(2)-AB))/(A+B)|=|(2sqrt(3)(a^(2)+b^(2)))/(-a(a^(2)+b^(2)))|`
` = sqrt(3) tan 60^(@) " " therefore theta = 60^(@)`.