Question 698944
<pre>
Ellipses like this {{{drawing(20,10,-2,2,-1,1,arc(0,0,-3.9,1.9) )}}} have equations of this form with aČ under the term in x 

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

Ellipses like this {{{drawing(10,20,-1,1,-2,2,arc(0,0,1.9,-3.9) )}}} have equations of this form with aČ under the term in y 

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

a is always greater than b, and of course aČ is greater than bČ.

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

The center is (h,k), the semi-major axis is a, the semi minor axis is b.

{{{(x-5)^2/81}}}{{{""+""}}}{{{(y-9)^2/25}}}{{{""=""}}}{{{1}}}

Since the denominator 81 is larger than the denominator 25, so we know
this is an ellipse that looks like this {{{drawing(20,10,-2,2,-1,1,arc(0,0,-3.9,1.9) )}}} 

By comparison h=5, k=9, aČ=81, bČ=25, and so a=9, b=5

So the center is (h,k) = (5,9). The major axis is 2a=18, and the minor
axis is 2b=10, and the two axes bisect each other at the center (5,9), 
so we draw this:

{{{drawing(400,3600/11,-6,16,-2,16, graph(400,3600/11,-6,16,-2,16),
green(line(-4,9,14,9),line(5,4,5,14)),locate(5,9,"(5,9)") )}}}{{{matrix(10,1,   
AND,THEN,WE,CAN,SKETCH,IN,THE,ELLIPSE,LIKE,"THIS:")}}}{{{drawing(400,3600/11,-6,16,-2,16, graph(400,3600/11,-6,16,-2,16),arc(5,9,18,-10),locate(5,9,"(5,9)"),
green(line(-4,9,14,9),line(5,4,5,14)) )}}}

Edwin</pre>