# uc3rmw_0802e04

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Hi there.
In this example, we want to find
the equation of a parabola given
a set of information about
its graph the problem states,
"Find the standard equation of the parabola
that satisfies the given conditions.
Also find the focal
diameter of each parabola."
For part A, we are given the focus 0
comma 5, the directrix y equals minus 3.
For part B we've given the vertex minus
1 comma 2, and directrix x equal 2.
And for part C we're given the vertex minus
5 comma minus 2, and focus 1 comma minus 2.
So first, let's recall the
general equation for a parabola.
The graph of Ax squared plus Cy
squared plus Dx plus Ey plus F
equals 0 is a parabola if AC equals 0, A
is not equal to 0, or C is not equal to 0.
That is, A is equal to 0, or C is
equal to 0, but not both are 0.
Let's recall that to find the
standard equation of a parabola we
must find the vertex h comma k, and p.
Remember that p is both the distance
between the vertex and the focus,
and the distance between the
vertex and the directrix.
The value 2p is the distance
between the focus and the directrix.
In part A, the distance between the focus
0 comma 5, and directrix y equal minus 3,
is 2p equals 5 minus in
parentheses minus 3, or 8 units.
We want to draw a rough sketch
to help us visualize this.
Now given that 2p is equal to 8, our vertex
is 1/2 times 8, or 4 units below the focus.
So p is equal to 4, and the vertex
is 0 comma 5 minus 4, or 0 comma 1.
We have that h comma k
is equal to 0 comma 1.
So h is equal to 0, and k is equal to 1.
So the standard equation, when
the directrix is below the vertex,
is x minus h squared equal
to 4p times y minus k.
And that becomes, given that h, k is equal
to 0 comma 1, x minus 0 squared equal
to 4 times p--
p is 4-- times y minus 1.
Now we can solve this for x squared.
We have that x squared is
equal to 16 times y minus 1.
And so the focal diameter is 4p,
or 4 times 4, which is equal to 16.
So now let's draw a rough graph to
visualize the graph of our parabola
for part A. We have our x-y axes here.
Given this form of the equation, we
have a parabola that opens upward,
where the vertex is 0 comma 1.
So we go 0 on the x-axis, 1 on the y-axis.
We locate our vertex and draw our parabola.
We have that the directrix
is y equal minus 3.
We'll put minus 3 at about that location,
and draw our directrix, y equals minus 3.
We can find our focal point--
h comma k plus p, which is 0
comma 1 plus 4, or 0 comma 5.
And so we can put that point
here on the y-axis at y equals 5.
So there's our focal point--
F 0 comma 5.
We also have the vertex
right here at 0 comma 1.
We'll go ahead and label that.
And so that's the rough sketch of the
graph of our equation from part A.
In part B, the distance between the vertex
minus 1 comma 2 in the directrix x equal 2
is p equal to 2 minus in
parentheses minus 1, or p equals 3.
We're going to draw a rough sketch
to help us visualize this as well.
The vertex, minus 1 comma 2, is equal to
h comma k, so that h is equal to minus 1,
and k is equal to 2.
The standard equation when the
directrix is to the right of the vertex
is y minus k squared equals
minus 4p times x minus h.
Now we can replace k with 2, and
h with minus 1, and also p with 3,
to obtain our equation.
We have y minus 2 squared equals minus
4 times p is equal to 3 times x minus h,
which is minus 1.
We simplify this a bit further.
We have y minus 2 squared
equals minus 12 times x plus 1.
So the focal diameter is 4p, which
is equal to 4 times 3, or 12.
Now we're going to sketch a rough graph
to help us visualize this equation.
Again, the directrix is to
the right of the vertex.
We have a parabola that opens left.
The vertex is at minus 1 comma 2.
So I'll place that right there
as the point minus 1 comma 2.
Well that is 2.
And that is minus 1.
Here's our x and y-axes labeled.
Our parabola opens to the left.
We have our directrix x equal 2.
And it's symmetric about
the line y equals 2.
The focal point is h minus p comma k,
or minus 1 minus 3 comma 2, or minus 4
comma 2.
So minus 4 is about here on the
x-axis, and 2 is right there.
So they meet up right at about that point.
So here's our focal point--
minus 4 comma 2.
So there's a rough sketch of what
our parabola looks like for part B.
So for part C, the distance between
the vertex minus 5 comma minus 2,
and the focus 1 comma
minus 2, is p equal to 1
minus in parentheses minus
5, which is equal to 6.
We're going to draw a
rough sketch of this one
as well to help us visualize the
equation that we're going to determine.
The vertex is minus 5 comma minus 2.
So our h is equal to minus 5.
Our k is equal to minus 2.
And so the equation with the
vertex to the left of the focus
is given by y minus k squared
equals 4 times p times x minus h.
You plug in that k is equal to
minus 2, and h is equal to minus 5.
And we plug-in that p is equal to 6.
We have y minus in parentheses minus
2 squared equal to 4 times 6 times
x minus h-- h being minus 5 in parentheses.
And we can simplify this a bit further.
We get the equation y plus 2
squared equals 24 times x plus 5.
So the focal diameter is 4 times
p, which is 4 times 6, or 24.
Now let's draw the rough graph of this
equation to visualize the situation.
We put our x-y axes down.
We have a parabola that opens to the right,
given this form of the equation, where
the vertex is minus 5 comma minus 2.
So we locate a minus 5 on our x-axis, and
a rough position for minus 2 on the y-axis.
The vertex we'll put right there.
Our parabola opens to the right with
the symmetry axis at y equal minus 2.
The directrix is the equation x
equal h minus p for the directrix--
h minus p being minus
5 minus 6, or minus 11.
So we'll go ahead and locate a
minus 11 roughly here on the x-axis
and draw our directrix.
So here is our directrix x equal minus 11.
We can also draw our focal
point for this equation.
Our focal point is the point h plus p
comma k, or minus 5 plus 6 comma minus 2.
And so the focal point is
reduced to 1 comma minus 2.
We locate 1, roughly on the
x-axis to be about there.
Minus 2 of course at that point.
So our focal point is labeled approximately
there, the point 1 comma minus 2.
So this is roughly what our
graph looks like for part C.