8)
[tex]\bf y^2+8y-8x=0\implies y^2+8y=8x[/tex]
now, let's group the left-side,
(y²+8y) = 8x
(y²+8y+?²) = 8x
we're kinda missing a fellow there, to get a perfect square trinomial, since in such trinomial the middle term is a product of 2 and the other terms, we can say [tex]\bf 2y\boxed{?}=8y\implies \boxed{?}=\cfrac{8y}{2y}\implies \boxed{?}=4[/tex]
so the missing fellow is just 4 then, however, recall, all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add 4², we also have to subtract 4²,
[tex]\bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\
p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------[/tex]
[tex]\bf y^2+8y-8x=0\implies y^2+8y=8x\implies (y^2+8y+4^2)-4^2=8x
\\\\\\
(y+4)^2=8x+16\implies (y+4)^2=8(x+2)
\\\\\\
(y+4)^2=4(2)(x+2)\implies [y-\stackrel{k}{(-4)}]^2=4(\stackrel{p}{2})[x-(\stackrel{h}{-2})][/tex]
so, the vertex is at (-2,-4).
leading term, namely y², has a positive coefficient, meaning that "p" is a positive value, which is just 2, and that the parabola is opening to the right-hand-side.
so, from the vertex -2,-4, we move over 2 units to the right to get the focus point, at (0, -4)
the directrix is "p" units in the opposite direction, so from -2,-4, we move to the left to get the directrix line of x = - 4.
the axis of symmetry comes from the vertex coordinates, the parabola is horizontal, thus the axis of symmetry comes from the y-coordinate, y = -4.
9)
now, here we also have a horizontal parabola, notice that the squared variable is the "y", thus, and without much fuss, we'll do the same as we did before, group it, "complete the square", and then set in "vertex and focus" form, to see what's "p" and h,k.
[tex]\bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\
p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------[/tex]
[tex]\bf 4x-y^2-4y=0\implies 4x=y^2+4y\implies 4x=(y^2+4y+\boxed{?}^2)
\\\\\\
4x=(y^2+4y+2^2)-2^2\implies 4x+4=(y^2+4y+2^2)
\\\\\\
4(x+1)=(y+2)^2\implies 4(\stackrel{p}{1})[x-\stackrel{h}{(-1)}]=[y-\stackrel{k}{(-2)}]^2[/tex]
since the leading term's coefficient is positive, "p" is positive as well, that also means the parabola is again, opening to the right-hand-side.
the vertex is clearly at (-1,-2).
the axis of symmetry is of course y = - 2
from the vertex we move 1 unit to the right to get the focus at (0, -2).
and we move in the opposite direction 1 unit to get the directrix at y = -2..
