Question 996230
 {{{ x^2 + 4y^2 - 16y = 0}}}


{{{ x^2 + 4(y^2 - 4y )= 0}}}........complete square

{{{ x^2/4 + (y^2 - 4y+2^2)-4= 0}}}

{{{ x^2/4 + (y- 2)^2-4= 0}}}

{{{ x^2/4 + (y- 2)^2=4}}}....both sides divide by {{{4}}}

{{{ x^2/16 + (y- 2)^2/4= 1}}}

{{{ x^2/16 +(y- 2)^2/4= 1}}}.....{{{ x^2/16}}} we can write as {{{ (x-0)^2/16}}} 


{{{ (x-0)^2/16 +(y- 2)^2/4= 1}}}

so, you have an ellipse that has:

{{{h=0}}},{{{k=2}}}
{{{a=4}}}=>semi-major axis length
{{{b=2}}}=>semi-minor axis length
{{{c^2=4^2-2^2}}}
{{{c^2=12}}}
{{{c=sqrt(12)}}}
{{{c=2sqrt(3)}}}

the center is at ({{{0}}},{{{2}}})
foci : ({{{c}}},{{{k}}}), ({{{-c}}},{{{k}}})
({{{2sqrt(3)}}}, {{{2}}})
({{{-2sqrt(3)}}}, {{{2}}})
vertices:  ({{{a}}}, {{{k}}}),  ({{{4-a}}}, {{{k}}})
 ({{{4}}}, {{{2}}})
({{{-4}}}, {{{2}}}) 

eccentricity : {{{c/a=2sqrt(3)/4=sqrt(3)/2}}} 

{{{ graph( 600, 600, -10, 10, -10, 10,sqrt(4(1-(x-0)^2/16) )+2,-sqrt(4(1-(x-0)^2/16) )+2) }}}