Question 1060958
{{{x^2+9y^2-2x+18y+1=0}}}
{{{x^2-2x+9y^2+18y=-1}}}
{{{x^2-2x+1+9y^2+18y+9=-1+1+9}}}
{{{(x-1)^2+9(y^2+2y+1)=9}}}
{{{(x-1)^2+9(y+1)^2=9}}}
You can continue to
{{{(x-1)^2/9+(y+1)^2=1}}} ,
but even before that last step, it is obvious that
{{{(x-1)^2<=9}}} while {{{(y+1)^2<=1}}} ,
so {{{x-1}}} ranges from {{{-3}}} to {{{3}}} ,
a difference of {{{highlight(6)}}} ,
and of course the extreme values of {{{x}}} also have a difference of {{{6}}} .
The length of the major axis is {{{highlight(6)}}} .
{{{drawing(600,300,-3,5,-3,1,grid(1),
green(line(1,-5,1,5)),green(line(-5,-1,5,-1)),
red(arc(1,-1,6,2,0,360)),red(circle(1,-1,0.05))
)}}}
 
NOTES:
The last two equations show that the ellipse is symmetrical with respect to {{{x=1}}} and {{{y=-1}}} , so (1,-1) is the center.
In general, an ellipse can be written in the form
{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}} , or {{{(x-h)^2/b^2+(y-k)^2/a^2=1}}} ,
and once you get to that form,
choosing {{{a}}} and {{{b}}} so that {{{a>b}}} ,
{{{a}}} is the semi-major axis, which is parallel to the x- or y-axis (whatever variable is on to of the {{{a^2}}} ),
{{{b}}} is the semi-minor axis,
{{{"( h , k )"}}} is the center,
and you can find the foci knowing that
they are on the major axis, at a distance {{{c}}} to each side of the center, with {{{a^2=b^2+c^2}}} .