The point (2,2) lies on the parabola `(x-3)^2=4a(y-1)` find the focus and the equation of the directrix.

The focus of the parabola x ^(2) =-16 y is

The focus of the parabola y ^(2) =4y -4x is

Focus of the parabola x^2-10x-4y+17=0 , is

Show that the equation of the parabola with vertex (2,5) and the equation of the directrix x=1 is (y-5)^2=-4(x-2)

The parabola x^2=4ay passes through (-2,1) Find the co-ordinates ends of the LR and equation of thr directrix.

The point P(a,b) lies on the straight line 3x+2y=13 and the point Q(b,a) lies on the straight line 4x-y=5 , then the equation of the line PQ is

The equation of the parabola with focus at (-6,0) and the line x+2=0 as directrix is

For the parabola 3y^2=16x , find the parameter of the point. (3,-4)