Question 1199474
<pre>
I forgot to do the foci above.

{{{drawing(320,420,-4,12,-9,15, 

graph(320,420,-4,12,-9,15),

graph(320,420,-4,12,-9,15,3+(5/4)sqrt(x^2-8x)),

graph(320,420,-4,12,-9,15,3-(5/4)sqrt(x^2-8x)),

blue(line(0,3,8,3), line(4,-2,4,8)), 

green(line(-5,-8.25,13,14.25),line(-5,14.25,13,-8.25)),

green(line(0,-2,8,-2), line(8,-2,8,8), line(8,8,0,8), line(0,8,0,-2)))}}}

The focal distance is c, which is found from the Pythagorean
relation {{{c^2=a^2+b^2}}}

{{{c^2=4^2+5^2}}}
{{{c^2=16+25}}}
{{{c^2=41}}}
{{{c=sqrt(41)}}}  <---(I think on your problem this comes out to be an integer.)

So the foci are {{{sqrt(41)}}} or about 6.4 units right and left of the
center. so we get two points right and left of the center about 6.4 units
from the center, the foci are approximately (4+6.4,3) and (4-6.4,3), which
are approximately the points (10.4,3) and (-2.4,3), as shown on the graph
below:

{{{drawing(320,420,-4,12,-9,15, 

graph(320,420,-4,12,-9,15),

graph(320,420,-4,12,-9,15,3+(5/4)sqrt(x^2-8x)),

graph(320,420,-4,12,-9,15,3-(5/4)sqrt(x^2-8x)),

blue(line(0,3,8,3), line(4,-2,4,8)),

circle(4-sqrt(41),3,.1), circle(4-sqrt(41),3,.15), circle(4-sqrt(41),3,.2),
circle(4+sqrt(41),3,.1), circle(4+sqrt(41),3,.15), circle(4+sqrt(41),3,.2),
green(line(-5,-8.25,13,14.25),line(-5,14.25,13,-8.25)),

green(line(0,-2,8,-2), line(8,-2,8,8), line(8,8,0,8), line(0,8,0,-2)))}}}



Edwin</pre>