Question 765278
{{{9x^2+36(y-2)^2-324=0}}}


{{{9x^2+36(y-2)^2=324}}}

{{{9x^2/324+36(y-2)^2/324=324/324}}}

{{{cross(9)x^2/cross(324)36+cross(36)(y-2)^2/cross(324)9=1}}}

{{{x^2/36+(y-2)^2/9=1}}} or

{{{(x-0)^2/36+(y-2)^2/9=1}}}

this is an ellipse with {{{h=0}}} and {{{k=2}}}; so, the center is at

({{{0}}},{{{2}}})

we also know that semi-major axis {{{a=6}}} and semi-minor axis {{{b=3}}}; so, we have an ellipse with horizontal major axis

vertices: 
({{{h-a}}},{{{k}}}) =>({{{0-6}}},{{{2}}})=>({{{-6}}},{{{2}}})
({{{h+a}}},{{{k}}})=>({{{0+6}}},{{{2}}})=>({{{6}}},{{{2}}})

c0-vertices:
({{{k}}},{{{k-b}}}) =>({{{2}}},{{{2-3}}})=>({{{2}}},{{{-1}}})
({{{k}}},{{{k+b}}})=>({{{2}}},{{{2+3}}})=>({{{2}}},{{{5}}})

foci:

The formula generally associated with the focus of an ellipse is

   {{{ c^2= a^2 - b^2}}} 

so, {{{b=3}}} and {{{a=6}}} 

{{{ c^2= 6^2 - 3^2}}}

{{{ c^2=36 - 9}}}

{{{ c^2=27}}}

{{{ c=sqrt(27)}}}

{{{ c=5.19}}} and {{{ c=-5.19}}}

so, foci is at  ({{{5.19}}},{{{2}}}) and ({{{-5.19}}},{{{2}}})


{{{ graph( 600, 600, -10,10, -10, 10,-sqrt(-x^2/4+9)+2 ,sqrt(-x^2/4+9)+2) }}}