Question 331126
Write an equation of a parabola whose vertex is at (-2,1) and focus is at (-3,1).
<pre><font size = 4 color = "indigo"><b>

We'll have to draw the parabola to determine which way it opens.

Let's draw the focus and the vertex:
{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5), 
line(-2+.07,1+.07,-2-.07,1-.07), line(-2+.07,1,-2-.07,1),line(-2,1+.07,-2,1-.07),
line(-2-.07,1+.07,-2+.07,1-.07),
locate(-2,.8,V),
line(-3+.07,1+.07,-3-.07,1-.07), line(-3+.07,1,-3-.07,1),line(-3,1+.07,-3,1-.07),
line(-3-.07,1+.07,-3+.07,1-.07),
locate(-3,.8,F)


 )}}}

The vertex and the focus are 1 unit apart. The vertex is halfway between the
focus and the directrix.  Since the vertex and the focus are 1 unit apart,
the directrix is the vertical line whose equation is x=-1, so we draw 
that directrix line in:

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5), 
line(-2+.07,1+.07,-2-.07,1-.07), line(-2+.07,1,-2-.07,1),line(-2,1+.07,-2,1-.07),
line(-2-.07,1+.07,-2+.07,1-.07),
locate(-2,.8,V),
line(-3+.07,1+.07,-3-.07,1-.07), line(-3+.07,1,-3-.07,1),line(-3,1+.07,-3,1-.07),
line(-3-.07,1+.07,-3+.07,1-.07),
locate(-3,.8,F), line(-1,-6,-1,6)


 )}}}

Now we draw a green line from the focus through the vertex to the
directrix: 

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5), 
line(-2+.07,1+.07,-2-.07,1-.07), line(-2+.07,1,-2-.07,1),line(-2,1+.07,-2,1-.07),
line(-2-.07,1+.07,-2+.07,1-.07),
locate(-2,.8,V),
line(-3+.07,1+.07,-3-.07,1-.07), line(-3+.07,1,-3-.07,1),line(-3,1+.07,-3,1-.07),
line(-3-.07,1+.07,-3+.07,1-.07),
locate(-3,.8,F), line(-1,-6,-1,6), green(line(-3,1,-1,1))


 )}}}

To help us sketch the parabola we construct two squares, 
one on each side of the green line from the vertex to the focus:

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5), 
line(-2+.07,1+.07,-2-.07,1-.07), line(-2+.07,1,-2-.07,1),line(-2,1+.07,-2,1-.07),
line(-2-.07,1+.07,-2+.07,1-.07),
locate(-2,.8,V),
line(-3+.07,1+.07,-3-.07,1-.07), line(-3+.07,1,-3-.07,1),line(-3,1+.07,-3,1-.07),
line(-3-.07,1+.07,-3+.07,1-.07),
locate(-3,.8,F), line(-1,-6,-1,6), green(line(-3,1,-1,1),
rectangle(-3,-1,-1,3))


 )}}}

Now we can sketch in the parabola with the vertex and which passes
through the corners of those two squares:

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5,1-2sqrt(-x-2)), graph(400,400,-5,5,-5,5,1+2sqrt(-x-2)), 
line(-2+.07,1+.07,-2-.07,1-.07), line(-2+.07,1,-2-.07,1),line(-2,1+.07,-2,1-.07),
line(-2-.07,1+.07,-2+.07,1-.07),
locate(-2,.8,V),
line(-3+.07,1+.07,-3-.07,1-.07), line(-3+.07,1,-3-.07,1),line(-3,1+.07,-3,1-.07),
line(-3-.07,1+.07,-3+.07,1-.07),
locate(-3,.8,F), line(-1,-6,-1,6), green(line(-3,1,-1,1),
rectangle(-3,-1,-1,3))
 )}}}

Remember the directrix is outside the parabola and the focus
is inside the parabola.

The "latus rectum" or "focal chord" is the line across the parabola
that goes through the focus and is parallel to the directrix. It's the
line consisting of the left sides of the two squares.  I'll draw it in
black below. Notice it is 4 units long:

{{{drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5,1-2sqrt(-x-2)), graph(400,400,-5,5,-5,5,1+2sqrt(-x-2)), 
line(-2+.07,1+.07,-2-.07,1-.07), line(-2+.07,1,-2-.07,1),line(-2,1+.07,-2,1-.07),
line(-2-.07,1+.07,-2+.07,1-.07),
locate(-2,.8,V),
line(-3+.07,1+.07,-3-.07,1-.07), line(-3+.07,1,-3-.07,1),line(-3,1+.07,-3,1-.07), 
line(-3-.07,1+.07,-3+.07,1-.07),
locate(-3,.8,F), line(-1,-6,-1,6), green(line(-3,1,-1,1),
rectangle(-3,-1,-1,3)), line(-2.99,-1,-2.99,3),line(-3.01,-1,-3.01,3)
 )}}}

The equation of the horizontal parabola which opens right or left and 
has vertex (h,k) is given by:

{{{(y - k)^2}}}{{{""=""}}}{{{"" +- m(x - h)}}}

where m is the length of the "latus rectum" or "focal chord", which
is the black line above, which we see is 4 units long, so m=4.

When a horizontal parabola opens right, the sign before m is +,
and when a horizontal parabola opens left, the sign before the
m is -, so here the sign is -.  The equation is:

{{{(y - 1)^2}}}{{{""=""}}}{{{-4(x + 2)}}} 

That's the equation of the parabola in standard form.

Edwin</pre>