For a parabola defined by the equation J:2 = y, determine the focus and directrix. (1 point) Find the equation of a parabola with its vertex at the origin and a focus of (8, 0). (1 point) y2 232x y2 =2x x2 =32y Rewrite the equation y2 2x+ 2y 5 = 0 in standard form. Determine the focus and directrix. (1 point) (yt)'=2(x+3) (y+1)'=2(x+3) p=[_2,_1],x=_3 2 2 _ y Use the image to answer the question. Write the standard form of the equation for the parabola shown in the graph. * (1 point) 2 011)2 =8(x+3) V'H'" 0F (1,1) 01)" = 2(x+3) -6 .4 .2 2 4 6 (2H3)2 = 80-1) -2 ("32201)