Question 386103
<pre>

{{{(x-1)^2/21 + (y-3)^2/4 = 1}}}

That is the form: 

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

because a is always greater than b in an ellipse.

It has center (h,k) = (1,3), semi-major axis = a = {{{sqrt(21)}}},
and semi-minor axis = b = {{{sqrt(4)=2}}}

It has its major axis horizontal because aČ is under the term in x

We plot the center:

{{{drawing(400,200, -10,10,-3,7,
locate(1,3,"(1,3)"),
graph(400,200,-10,10,-3,7), circle(1,3,.1) )}}}

We draw the complete major axis {{{a = sqrt(21)}}}, about 4.6 units right and
left from the center (in green).  The endpoints of the major axes are the
vertices. Since the vertices are {{{sqrt(21)}}} units right and left of the
center, their coordinates are (1 ± {{{sqrt(21)}}},3)

{{{drawing(400,200, -10,10,-3,7,

graph(400,200,-10,10,-3,7), circle(1,3,.1),
locate(1+sqrt(21),3,V(1+sqrt(21),3)), locate(-8,3,V(1-sqrt(21),3)),

green(line(1-sqrt(21),3,1+sqrt(21),3))
 )}}}

We draw the complete minor axis {{{b = 2}}} units above and below
the center (also in green):
 
{{{drawing(400,200, -10,10,-3,7,

graph(400,200,-10,10,-3,7), circle(1,3,.1),
locate(1+sqrt(21),3,V(1+sqrt(21),3)), locate(-8,3,V(1-sqrt(21),3)),

green(line(1-sqrt(21),3,1+sqrt(21),3), line(1,5,1,1))
 )}}}

We sketch in the ellipse:
{{{drawing(400,200, -10,10,-3,7,

graph(400,200,-10,10,-3,7), circle(1,3,.1),
locate(1+sqrt(21),3,V(1+sqrt(21),3)), locate(-8,3,V(1-sqrt(21),3)),
arc(1,3,2*sqrt(21),-4),
green(line(1-sqrt(21),3,1+sqrt(21),3), line(1,5,1,1))
 )}}}

Finally we find the foci which are c units from the center and are
located on the major axis, where 

{{{c = sqrt(a^2-b^2)=sqrt(21-4)=sqrt(17)}}}, about 4.1 units.

So the coordinates of the foci are (1 ± {{{sqrt(17)}}},3)
  
I'll plot them.  They are just inside the ellipse on each end:

{{{drawing(400,200, -10,10,-3,7,

graph(400,200,-10,10,-3,7), circle(1,3,.1),
locate(1+sqrt(21),3,V(1+sqrt(21),3)), locate(-8,3,V(1-sqrt(21),3)),
arc(1,3,2*sqrt(21),-4),
green(line(1-sqrt(21),3,1+sqrt(21),3), line(1,5,1,1)),
circle(1+sqrt(17),3,.15), circle(1-sqrt(17),3,.15)


 )}}}

Edwin</pre>