# y= 1/4 (x^2-2x+5) = 1/4 (x^2-2x +1) -1/4+5/4 = 1/4(x-1)^2+1#
Comparing with standard equation # y= a(x-h)^2+k : (h,k)# being vertex we find here vertex at #h=1 , k=1 or (1,1) ; a=1/4#
Vertex is at #(1,1)#
We know vertex #(1,1)# is at equidistance from focus and directrix and are at opposite side of vertex. We also know #a= 1/(4d) or d = 1/(4|a|)=1/(4*1/4)=1# , where #d# is the distance of directrix from vertex.
Here #a >0# so parabola opens upwards. so directrix is #y=(1-1) or y= 0 #
Focous is above the vertex at a distance of #4# unit. so focus is at #1,(1+1) or (1,2)# graph{1/4(x^2-2x+5) [-10, 10, -5, 5]} [Ans]
