SOLUTION: 1. Find the Equation for the parabola with focus (3,2) and directrix y=6. 2. Write the equation for the circle with center (3,-4) and radius 5square root 2. 3. write the equa

1. Find the Equation for the parabola with focus
  (3,2) and directrix y=6. 

The focus is a point and the directrix is a line.

The parabola has the equation

     (x - h)² = 4p(y - k)

where (h,k) is the vertex, p is the distance from
the vertex to the focus and also the distance
from the vertex to the focus. 

Lets draw the focus F(3,2). I'll put an "o" there.
And I'll draw the directrix y=6 which is a horizontal
line 6 units above the x-axis.  I'll draw it green



The vertex is halfway between the focus and the
directrix.  So the vertex is at the point V(3,4).

I'll plot the vertex with an o.



That means that the vertex (h,k) = (3,4)

So in the equation 

     (x - h)² = 4p(y - k)

we can now put in h and k:

     (x - 3)² = 4p(y - 4)

Now we only need p, which is the
distance from the vertex to the focus 
and also the distance from the
vertex to the directrix. 

So the distance from V(3,4) to F(3,2)
is 2 units, so p = 2.

In the equation

     (x - 3)² = 4p(y - 4)     

we can put in 2 for p and have

     (x - 3)² = 4(2)(y - 4)

or

     (x - 3)² = 8(y - 4)

That's the equation, but now we have to
draw the graph. 

Draw a line from the focus thru
the vertex to the directrix.



Now draw a square on the left of that line.



Now draw a square on the right of that line:



Finally sketch in the parabola going through the vertex
and the corners or the two squares:



That's the graph.

----------------------------------------

2. Write the equation for the circle with center (3,-4) 
               _ 
   and radius Ö5

   (x - h)² + (y - k)² = r²

is the equation of the circle with center (h,k) and radius r.  So we
substitute (h,k) = (3,-4) 
                               _
   (x - 3)² + (y - (-4) )² = (Ö5)²

   (x - 3)² + (y + 4)² = 5


   
------------------------------------------------------------

4. Write equation for the ellipse with foci (9,0) and (-9,0) and endpoints on the verticle axis of (0,2) and (0,-2).

The equation of such an ellipse is

     (x - h)²    (y - k)²
     -------- + ---------- = 1
        b²          a²

where (h,k) is the center
a = half the length of the major axis
and b = half the minor axis
and a² = b² + c² where c is the distance from
Center to foCus.  (You can remember that because
the words "Center" and "foCus" both have the 
letter C, and "vertex" doesn't.)

Lets plot the foci F(-9,0) and F(9,0) and the 
ends of the minor (vertical) axes (0,-2) and
(0,2)



  
Now we know that the center is halfway between the
foci, and the origin (0,0) is halfway between the
two foci, so we know that (h,k) = (0,0).

We know that c is the distance from the Center to
a foCus, which is 9 units, so c = 9 

We also know that b = half the minor (vertical in this 
case) axis.  Since the minor axis is the line from (0,-2)
to (0,2), and it is 4 units long, then half the
minor axis is 2, and so b = 2, so we can fill those
into:

      (x - h)²    (y - k)²
     -------- + ---------- = 1
        b²          a²

and get

     (x - 0)²    (y - 0)²
     -------- + ---------- = 1
        2²          a²

or

      x²    y²
     --- + ---- = 1
      4     a²

So all we need is a, which is half
the length of the major axis.  We
can calculate a from the relation

   a² = b² + c²
   a² = 2² + 9²
   a² = 4 + 81
   a² = 85

         __
    a = Ö85 = 9.2 approximately

This puts the vertices at V(-9.2,0) and V(9.2,0) and they
are very close to the focal points:
 


Now that we know that a² = 85, the equation of the
ellipse is:

      x²    y²
     --- + ---- = 1
      4     85

and we can sketch in the ellipse: 

   graph(800,266.7,-12,12,-4,4)     )}}}



Edwin