How do you identify the conic section represented by #(y-4)^2= 8(x-1)# and what are the critical points?

1 Answer
Apr 25, 2018

Answer:

It is a horizontal parabola and does not have any critical point.

Explanation:

A conic section of the form #(y-k)^2=a(x-h)# represents a horizontal parabola, whose vertex is #(h,k)# and axis of symmetry is #y-k=0# or #y=k#

Hence for #(y-4)^2=8(x-1)#, vertex is #(1,4)# and axis of symmetry is #y=4#.

Critical points are those points at which #(dy)/(dx)=0#

As #(y-4)^2=8(x-1)#, implicit differentiation gives

#2(y-4)(dy)/(dx)=8# or #(dy)/(dx)=4/(y-4)=4/sqrt(8(x-1))#

As such, #(dy)/(dx)# is never #0# and we do not have a critical point.

For a vertical parabola, it is #(x-h)^2=a(y-k)#, whose vertex is #(h,k)# and axis of symmetry is #x=h#. in this case we will have a critical point at #(h,k)#.

graph{(y-4)^2=8(x-1) [-8.09, 11.91, -1.28, 8.72]}