Answer
$ a.\qquad$
New vertex:$\quad V'(1,-2)$
New focus: $ \quad F'(3,-2)$
New directrix: $ \quad x=-1$
$ b.\qquad$
See graph.
Work Step by Step
$ a.\qquad$
$y^{2}=8x=4(2)x$
has the form
$y^{2}=4px $ ,$\qquad x=\displaystyle \frac{y^{2}}{4p}$ ,$\qquad$... opens right
Vertex = $(0,0)$
Focus = $(p,0)=(2,0)$
Directrix: $x=-p=-2$
New vertex:
$\left\{\begin{array}{ll}
x'=x+1=0+1=1 & \text{...shift right}\\
y'=y-2=0-2=-2 & \text{...shift down}
\end{array}\right., \quad V'(1,-2)$
New focus:
$\left\{\begin{array}{ll}
x'=x+1=2+1=3 & \text{...shift right}\\
y'=y-2=0-2=-2 & \text{...shift down}
\end{array}\right., \quad F'(3,-2)$
New directrix (vertical line, shifted to the right): $x=-1$
$ b.\qquad$
See graph
The vertex and focus are now on the line $y=-2$
(the x-axis is shifted down).
