Answer
(a)
Vertex: $V(2,-5)$
Focus: $F(\frac{1}{2},-5)$
Directrix: $x=\frac{7}{2}$
(b)
Work Step by Step
Equation of a parabola with horizontal axis and vertex at $(h,k)$:
$(y-k)^2=4p(x-h)$
$(y+5)^2=-6x+12$
$(y+5)^2=-6(x-2)$
$[y-(-5)]^2=-6(x-2)$
$h=2$
$k=-5$
Vertex: $V(2,-5)$
$4p=-6$
$p=-\frac{3}{2}$. Parabola opens to the left.
The given equation can be obtained by shifting
$y^2=-6x$
right 2 units and downward 5 units. In this equation:
Focus: $F(p,0)=F(-\frac{3}{2},0)$
Directrix: $x=-p=\frac{3}{2}$
Now, shift the vertex and the directrix 2 units to the right and 5 units downward:
Focus: $F(\frac{1}{2},-5)$
Directrix: $x=\frac{7}{2}$
