The parabola whose equation is 4y^2+3x=2+8y, determine the co-ordinates of the focus, equation of the directrix and the length of the latus rectum?
recall the properties of the parabola
y^2 = 4px
Your parabola has been shifted so the vertex is at (2,-1) because
4y^2+3x=2+8y
4y^2+8y = -3x+2
4(y^2+2y+1) = -3x+2+4
4(y+1)^2 = -3x+6 = -3(x-2)
(y+1)^2 = (-3/4)(x-2)
Now just apply the properties of the standard parabola.