SOLUTION: Is this correct: What is the equation of a parabola with vertex (0,0) and focus (0,4)? {{{(x-h)^2 = 4p(y-k)}}} {{{(x-0)^2 = 4(4)(y-0)}}} {{{(x-0)^2 = 16(y-0)}}} If not wher

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Is this correct: What is the equation of a parabola with vertex (0,0) and focus (0,4)? {{{(x-h)^2 = 4p(y-k)}}} {{{(x-0)^2 = 4(4)(y-0)}}} {{{(x-0)^2 = 16(y-0)}}} If not wher      Log On


   



Question 194756: Is this correct: What is the equation of a parabola with vertex (0,0) and focus (0,4)?
%28x-h%29%5E2+=+4p%28y-k%29
%28x-0%29%5E2+=+4%284%29%28y-0%29
%28x-0%29%5E2+=+16%28y-0%29
If not where did I go wrong?

Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You are spot on, except that I would have gotten rid of the zeros in the final form.



John


Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
Edwin's solution and graph:

I think you're right, except maybe you'd want to erase
the 0's in the final answer.

I'll go through it.

Draw the two points, vertex V(0,0) and the focus F(0,4)



We can tell the parabola opens upward since the focus
is above the vertex.  The equation is 

%28x-h%29%5E2+=+4p%28y-k%29

Where the vertex is (h,k), and p is the directed
distance from the vertex to the focus.

If we put our pencil on the vertex and go to the focus
we move the pencil upward 4 units, so the parabolic constant
p = +4.  (When the pencil moves downward in going from the
vertex to the focus, then p is negative)

The directed distance from the vertex to the focus in
this problem is +4, and (h,k) = (0,0) so, substituting

%28x-h%29%5E2+=+4p%28y-k%29

%28x-0%29%5E2+=+4%284%29%28y-0%29

x%5E2+=+16y

To draw the finished parabola, first draw the directrix
which is a horizontal line p units on the other side of the 
vertex.  In this case it is 4 units below the vertex.  It
has the eqution y=-4



Now draw a square with its left side as the line from the directrix
to the focus:



Now draw another square with its right side as the line from the directrix
to the focus:



Finally sketch in the parabola passing through the 
upper far corners of the squares and through the vertex:



Edwin