Question 830525
<pre>
{{{4y^2-4y-4x+5}}}{{{""=""}}}{{{"0"}}}

You want it to look like this

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

which is a parabola with a horizontal axis of symmetry that
either opens left or right. We can't tell which yet.  It
has a vertex of (h,k)

{{{4y^2-4y-4x+5}}}{{{""=""}}}{{{"0"}}}

Get the y-terms on the left and everything else on the right

{{{4y^2-4y}}}{{{""=""}}}{{{4x-5}}}

Divide every term by 4

{{{y^2-y}}}{{{""=""}}}{{{x-5/4}}}

Complete the square on the left side:

1. To the side, multiply the coefficient of y, which is -1, by {{{1/2}}},
   getting {{{-1/2}}}
2. Square the result of 1.  {{{(-1/2)^2=1/4}}}
3. Add the result of 2 to both sides of the equation:

{{{y^2-y+1/4}}}{{{""=""}}}{{{x-5/4+1/4}}}

Factor the left side:  {{{(y-1/2)(y-1/2)=(y-1/2)^2}}}
Combine the numbers on the right {{{-5/4+1/4=-4/4=-1}}}

{{{(y-1/2)^2}}}{{{""=""}}}{{{x-1}}}

To show the 4p in the standard equation, perhaps your teacher
wants you to put a 1 factor on the right side:

{{{(y-1/2)^2}}}{{{""=""}}}{{{1(x-1)}}}

and now it corresponds exactly to

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

The vertex is (h,k) = (1,{{{1/2}}})

4p=1, so p={{{1/4}}}, since p is positive it opens right.

Its focus is the point {{{1/4}}} unit right of its vertex,
at (1,{{{3/4}}}), and the latus rectum is 4p=1 unit long
through the focus.  The directrix line is the vertical 
line {{{1/4}}} unit left of the vertex.  It has the 
equation {{{x=3/4}}}. to 4p = 1 unit. So we draw the 
vertex, focus, directrix and latus rectum and we have this:

{{{drawing(400,400,-4,4,-4,4,

graph(400,400,-4,4,-4,4),

circle(1,0.5,0.05),circle(1,0.5,0.03),circle(1,0.5,0.01),

circle(1.25,0.5,0.05),circle(1.25,0.5,0.03),circle(1.25,0.5,0.01),

green(line(3/4,-20,3/4,20),line(1.25,0,1.25,1))



 )}}}

Then we sketch in the parabola:

{{{drawing(400,400,-4,4,-4,4,

graph(400,400,-4,4,-4,4,1/2+sqrt(x-1)),
red(arc(1.4,1/2,.9,-.9,150,230)),
graph(400,400,-4,4,-4,4,1/2-sqrt(x-1)),

circle(1,0.5,0.07),circle(1,0.5,0.05),circle(1,0.5,0.03),circle(1,0.5,0.01),

circle(1.25,0.5,0.07),circle(1.25,0.5,0.05),circle(1.25,0.5,0.03),circle(1.25,0.5,0.01),

green(line(3/4,-20,3/4,20),line(1.25,0,1.25,1))




  )}}}

Yes, I know you didn't need to graph it but you'll have to later.

Edwin</pre>