Question 79604
<pre><font face = "lucida console"><size = 3><b>

draw the parabola. identify the focus and directrix.

4y = 7x^2 

You will need to get the equation in this form:

x² = 4py

where the vertex is (0,0) and the distance from the vertex to
focus is p and the distance from the vertex to the directrix
is -p.  [A distance is upward if it's positive and downward if
it's negative].  It's focal chord (or "latus rectum"), which
is the distance across the parabola at the focal point, has
length 4p.

      4y = 7x²

Write that as

     7x² = 4y

Divide both sides by 7

      x² = {{{4/7}}}y

Compare that to
  
      x² = 4py

So the vertex is (0,0) and 4p = 4/7, or p = 1/7,
so the focus is +1/7 units (above) the vertex or
the point (0,1/7). The vertex is the origin and
the focus is the point (0, 1/7) marked with an X
below:

{{{graph(400,400, -1.2,1.2,-1.2,1.2,(-x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),(x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x))) }}}

The directrix is -1/7 units (below) the vertex, which is a horizontal
line whose equation is y = -1/7.  Here is the directrix:

{{{graph(400,400, -1.2,1.2,-1.2,1.2,(-x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),(x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),-1/7)}}}
  
Next we draw the focal chord (or "latus rectum") which is 4p units or 4/7
units long which means it is 2/7 units on each side of the focus: 

{{{graph(400,400, -1.2,1.2,-1.2,1.2,(-x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),(x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),-1/7,(1/7)( sqrt(x+2/7)/sqrt(x+2/7) )( sqrt(2/7-x)/sqrt(2/7-x))    )}}}

And finally we can sketch in the parabola, whose vertex is the origin, and
which just touches the ends of the focal chord (or "latus rectum"):

{{{graph(400,400, -1.2,1.2,-1.2,1.2,(-x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),(x+1/7)(sqrt(x+.08)/sqrt(x+.08))(sqrt(.08-x)/sqrt(.08-x)),-1/7,(1/7)( sqrt(x+2/7)/sqrt(x+2/7) )( sqrt(2/7-x)/sqrt(2/7-x)),(7/4)x^2    )}}}

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

write the standard form of the equation of the circle that passes through
(1, -3) and whose center is the origin. 

The standard form of a circle with center at the origin is 

      x² + y² = r²

where the radius is r.

Since it passes through (x,y) = (1,-3), we substitute:

        (1)² + (-3)² = r²

               1 + 9 = r²

                  10 = r²

So the standard equation is 

             x² + y² = 10

So its center is the origin and its radius is {{{sqrt(10)}}}
or about 3.2. So we can sketch the graph with a compass.
Its graph is:

{{{graph(300,300,-4,4,-4,4,sqrt(10-x^2),-sqrt(10-x^2))}}} 
 
---------------------------------

draw the circle 3x² + 3y² = 48.

We divide through by 3 and get

                x² + y² = 16

Compare to 
                x² + y² = r²

and we see that r² = 16 and r, the radius is 4,

so the graph of this circle is

{{{graph(375,375,-5,5,-5,5,sqrt(16-x^2),-sqrt(16-x^2))}}}

Edwin</pre>