Schniz said:
sketch the graph x=y^2 marking its focus on the graph.
i am sure this is easy but i cant work out how to find the focus.
thanks.
y^2 = 4ax, where 4a = 1 , a = 1/4
SO, its a normal parabola, except that its rotated 90 degrees clockwise, and the focus is (1/4, 0), directrix is x=-1/4, vertex remains at (0,0).
hope it helps, M.D.