Write the equation in standard parabola form?
x^2+ 4xy +4y^2 +5root5 y+ 5 = 0
Thanks so much!
tan2θ = B/(A-C) = -4/3
Now, 2θ is in QII or QIV
If in QII,
cos2θ = -3/5
sinθ = √((1+3/5)/2) = 2/√5
cosθ = 1/√5
That gives us
x = 1/√5 (x' - 2y')
y = 1/√5 (2x' + y')
x^2 = 1/5 (x'^2 - 4x'y' + 4y'^2)
xy = 1/5 (2x'^2 - 3x'y' - 2y'^2)
y^2 = 1/5 (4x'^2 + 4x'y' + y'^2)
x^2+4xy+4y'^2 = 5x'^2
5√5y = 5(2x'+y')
So, adding it up we get
5x'^2 + 10x' + 5y' + 5 = 0
(x'+1)^2 + y'+1 = 0