Question 979639
find an equation of the ellipse having a major axis of length 12 and foci at
(1,2) and (-3,2)
<pre>We plot the foci (1,2) and (-3,2). 

Then we plot the center which is the midpoint between the two foci,
which is (-1,2). Then we draw the major axis 12 units long,
with the center at the middle, which means that we draw it 6 units 
on each side of the focus (in green).  That means that the vertices are 
(-7,2) and (5,2). 

{{{drawing(400,400,-9,7,-6,10,graph(400,400,-9,7,-6,10),
green(line(-7,2,5,2)),


circle(-3,2,0.15),circle(-3,2,0.13),circle(-3,2,0.11),circle(-3,2,0.09),circle(-3,2,0.07),circle(-3,2,0.05),circle(-3,2,0.03),circle(-3,2,0.01),

circle(-1,2,0.15),circle(-1,2,0.13),circle(-1,2,0.11),circle(-1,2,0.09),circle(-1,2,0.07),circle(-1,2,0.05),circle(-1,2,0.03),circle(-1,2,0.01),

circle(1,2,0.15),circle(1,2,0.13),circle(1,2,0.11),circle(1,2,0.09),circle(1,2,0.07),circle(1,2,0.05),circle(1,2,0.03),circle(1,2,0.01),

circle(-7,2,0.15),circle(-7,2,0.13),circle(-7,2,0.11),circle(-7,2,0.09),circle(-7,2,0.07),circle(-7,2,0.05),circle(-7,2,0.03),circle(-7,2,0.01),

circle(5,2,0.15),circle(5,2,0.13),circle(5,2,0.11),circle(5,2,0.09),circle(5,2,0.07),circle(5,2,0.05),circle(5,2,0.03),circle(5,2,0.01) )}}}

We know that this ellipse looks like this:{{{drawing(20,10,-2,2,-1,1,arc(0,0,-3.9,1.9) )}}}.

Therefore its equation is 

{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}

The center is (h,k) = (-1,2).
We know that "a" = half the major axis = half of 12 = 6

So we can fill in h,k, and a:

{{{(x+1)^2/6^2+(y-2)^2/b^2=1}}}

{{{(x+1)^2/36+(y-2)^2/b^2=1}}}
We only need b, the semi-minor axis:

{{{drawing(400,400,-9,7,-6,10,graph(400,400,-9,7,-6,10),
green(line(-7,2,5,2)),

arc(-1,2,12,8sqrt(2)),

circle(-3,2,0.15),circle(-3,2,0.13),circle(-3,2,0.11),circle(-3,2,0.09),circle(-3,2,0.07),circle(-3,2,0.05),circle(-3,2,0.03),circle(-3,2,0.01),

circle(-1,2,0.15),circle(-1,2,0.13),circle(-1,2,0.11),circle(-1,2,0.09),circle(-1,2,0.07),circle(-1,2,0.05),circle(-1,2,0.03),circle(-1,2,0.01),

circle(1,2,0.15),circle(1,2,0.13),circle(1,2,0.11),circle(1,2,0.09),circle(1,2,0.07),circle(1,2,0.05),circle(1,2,0.03),circle(1,2,0.01),

circle(-7,2,0.15),circle(-7,2,0.13),circle(-7,2,0.11),circle(-7,2,0.09),circle(-7,2,0.07),circle(-7,2,0.05),circle(-7,2,0.03),circle(-7,2,0.01),

circle(5,2,0.15),circle(5,2,0.13),circle(5,2,0.11),circle(5,2,0.09),circle(5,2,0.07),circle(5,2,0.05),circle(5,2,0.03),circle(5,2,0.01) )}}}

We calculate b from {{{c^2=a^2-b^2}}}
                    {{{2^2=6^2-b^2}}}
                    {{{4=36-b^2}}}
                    {{{b^2=32}}}
                    {{{b=sqrt(32)}}}
                    {{{b=sqrt(16*2)}}}
                    {{{b=4sqrt(2)}}}, approximately 5.7 units
                    up and down from the center. 

So the equation is 

{{{(x+1)^2/36+(y-2)^2/32=1}}}  
                   
It looks like a circle but I'll draw a circle (in red) so you
can see that it's not quite a circle:

{{{drawing(400,400,-9,7,-6,10,graph(400,400,-9,7,-6,10),
green(line(-7,2,5,2)),
red(circle(-1,2,6)),
arc(-1,2,12,8sqrt(2)),

circle(-3,2,0.15),circle(-3,2,0.13),circle(-3,2,0.11),circle(-3,2,0.09),circle(-3,2,0.07),circle(-3,2,0.05),circle(-3,2,0.03),circle(-3,2,0.01),

circle(-1,2,0.15),circle(-1,2,0.13),circle(-1,2,0.11),circle(-1,2,0.09),circle(-1,2,0.07),circle(-1,2,0.05),circle(-1,2,0.03),circle(-1,2,0.01),

circle(1,2,0.15),circle(1,2,0.13),circle(1,2,0.11),circle(1,2,0.09),circle(1,2,0.07),circle(1,2,0.05),circle(1,2,0.03),circle(1,2,0.01),

circle(-7,2,0.15),circle(-7,2,0.13),circle(-7,2,0.11),circle(-7,2,0.09),circle(-7,2,0.07),circle(-7,2,0.05),circle(-7,2,0.03),circle(-7,2,0.01),

circle(5,2,0.15),circle(5,2,0.13),circle(5,2,0.11),circle(5,2,0.09),circle(5,2,0.07),circle(5,2,0.05),circle(5,2,0.03),circle(5,2,0.01) )}}}
 
Edwin</pre>