Question 141028
What is the vertex and focus of the parabola whose equation is

{{{(y-8)^2 = -4(x-4)}}}?
<pre><font size = 4 color = "indigo"><b>
You must memorize:

"U-parabolas"

Parabolas which open upward or downward have equation:

{{{(x - h)^2 = 4p(y-k)}}}

with vertex ({{{h}}},{{{k}}}), focus ({{{h}}},{{{k+p}}}),

the horizontal line which is the directrix has the
equation {{{y=k-p}}}.

the vertical line which is the axis of symmetry has the
equation {{{x=h}}}.
 
If p is positive the parabola opens upward. If p is
negative the parabola opens downward

-----

C-parabolas

Parabolas which open rightward or leftward have equation:

{{{(y - k)^2 = 4p(x-h)}}}

with vertex ({{{h}}},{{{k}}}), focus ({{{h+p}}},{{{k}}}),

the vertical line which is the directrix has the
equation {{{x=h-p}}}.

the horizontal line which is the axis of symmetry has the
equation {{{y=k}}}.

If p is positive the parabola opens rightward. If p is
negative the parabola opens leftward.

----

{{{(y-8)^2 = -4(x-4)}}}

we compare this to

{{{(y-k)^2 = 4p(x-h)}}}, so the parabola opens right or left

{{{-h = -4}}} so {{{h = 4}}}

{{{-k = -8}}} so {{{k = 8}}}

{{{4p = -4}}} so {{{p = -1}}}, a negative number, so the
parabola opens left

So the vertex is ({{{h}}},{{{k}}}) = (4,8)

The focus is ({{{h+p}}}, {{{k}}}) = ({{{4+(-1)}}},{{{8}}}) = ({{{4-1}}},{{{8}}}) = ({{{3}}},{{{8}}})

the vertical line which is the directrix has the
equation {{{x=h-p}}}, or {{{x=4-(-1)}}} or {{{x=4+1}}} or {{{x=5}}}

the vertical line which is the axis of symmetry has the
equation {{{y=k}}}, or {{{y=8}}}.

To draw the parabola,  plot the focus, vertex, and directrix

{{{drawing(260,400,-7,7,-2,16 , rectangle(5,-10,10,20),
graph(260,400,-7,7,-2,16), 
locate(4-.2,8+.4,o), locate(3-.2,8+.4,o) )}}}

Draw a line from the focus thru to vertex to the directrix.

{{{drawing(260,400,-7,7,-2,16 ,

   graph(260,400,-7,7,-2,16), rectangle(5,-10,10,20),
locate(4-.2,8+.4,o), locate(3-.2,8+.4,o), 
line(3,8,5,8) 
 )}}}

Use that line as a side of a square, draw one square above:

{{{drawing(260,400,-7,7,-2,16 ,

   graph(260,400,-7,7,-2,16), rectangle(5,-10,10,20),
locate(4-.2,8+.4,o), locate(3-.2,8+.4,o), 
line(3,8,5,8),rectangle(3,8,5,10) 
 )}}}

and draw another square below:

{{{drawing(260,400,-7,7,-2,16 ,

   graph(260,400,-7,7,-2,16), rectangle(5,-10,10,20),
locate(4-.2,8+.4,o), locate(3-.2,8+.4,o), 
line(3,8,5,8),rectangle(3,8,5,10), rectangle(3,8,5,6) 
 )}}}

Finally sketch in the parabola with the vertex ({{{4}}},{{{8}}})
passing through corners of those two squares.

{{{drawing(260,400,-7,7,-2,16 ,

   graph(260,400,-7,7,-2,16,8+sqrt(-4(x-4))),

   graph(260,400,-7,7,-2,16,8-sqrt(-4(x-4))), rectangle(5,-10,10,20),
locate(4-.2,8+.4,o), locate(3-.2,8+.4,o), rectangle(3,8,5,10),
rectangle(3,8,5,6)
 
 )}}}

Edwin</pre>