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

Apr 25, 2018

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

Explanation:

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

Hence for ${\left(y - 4\right)}^{2} = 8 \left(x - 1\right)$, vertex is $\left(1 , 4\right)$ and axis of symmetry is $y = 4$.

Critical points are those points at which $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

As ${\left(y - 4\right)}^{2} = 8 \left(x - 1\right)$, implicit differentiation gives

$2 \left(y - 4\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 8$ or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{y - 4} = \frac{4}{\sqrt{8 \left(x - 1\right)}}$

As such, $\frac{\mathrm{dy}}{\mathrm{dx}}$ is never $0$ and we do not have a critical point.

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

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