Question 1167661
<pre>
Instead of doing your problem for you, I'll do one exactly like yours step-by-
step.  This is the problem I will do for you.  Use it as a model to do yours by:
</pre>Write the standard equation of the parabola given the two endpoints of
the focal width ; (-4,5) and (9,5) with p>0<pre>

The equation has the standard form:

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

where
the focus is the midpoint between the endpoints of the 
focal chord (also called the latus rectum)
4p = the focal width = distance between the two endpoints
p = the distance from the focus to the vertex 
(h,k) = the vertex, which has the same x-coordinate as the focus.

the focus is the midpoint between (-4,5) and (9,5) which is 
{{{matrix(1,3,
(matrix(1,3,(-4+9)/2,",",(5+5)/2)),
""="",
(matrix(1,3,5/2,",",5)) )}}}

I'll plot the end points of the focal width, and the focus:

{{{drawing(1.2*400,1.2*6000/31,-13,18,-5,10,circle(-4,5,.3),circle(9,5,.3),circle(5/2,5,.3),
graph(1.2*400,1.2*6000/31,-13,18,-5,10) )}}}

4p = the focal width = distance between the two endpoints, 
{{{matrix(1,9,4p,""="",
sqrt((9-(-4))^2+(5-5)^2),
""="",
sqrt((9+4)^2+(0)^2),
""="",
sqrt(13^2),
""="",
13)}}}

{{{4p=13}}}
{{{p=13/4}}}

p = the distance from the focus to the vertex.

So we subtract 13/4 from the y-coordinate of the focus:
The y-coordinate of the focus is 5, so
{{{5-13/4=20/4-13/4=7/4}}}

So the vertex is

{{{(matrix(1,3,5/2,",",7/4))}}}

I'll plot the vertex:

{{{drawing(1.2*400,1.2*6000/31,-13,18,-5,10,circle(-4,5,.3),circle(9,5,.3),circle(5/2,5,.3), circle(5/2,7/4,.3),
graph(1.2*400,1.2*6000/31,-13,18,-5,10) )}}}

and sketch in the parabola:

{{{drawing(1.2*400,1.2*6000/31,-13,18,-5,10,circle(-4,5,.3),circle(9,5,.3),circle(5/2,5,.3), circle(5/2,7/4,.3),
graph(1.2*400,1.2*6000/31,-13,18,-5,10,(1/13)(x^2-5x+29)) )}}}

So the standard equation of the parabola is

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

{{{(x-5/2)^2=13(y-7/4)}}}

Now do yours exactly step-by-step the same way.

Edwin</pre>