Question

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

Answer 1

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]}