Question 768228

{{{x^2 + 4y^2 + 12x - 32 y +96 = 0}}}.....rearrange terms

{{{x^2+ 12x + 4y^2  - 32 y +96=0}}}

{{{(x^2 + 12x+_ )+ (4y^2  - 32y +_)+96=0}}}.......complete the square, write {{{96}}} as 
{{{100-4}}} then write {{{100}}} as {{{36+64}}}

{{{x^2 + 12x+36 + 4y^2  - 32y +64-4=0}}}....group

{{{(x^2 + 12x+36 )+ 4(y^2  - 8y +16)-4=0}}}

{{{(x+6 )^2+ 4(y  - 4)^2-4=0}}}

{{{(x+6 )^2+ 4(y  - 4)^2=4}}}.......both sides divide by {{{4}}}

{{{(x+6 )^2/4+ 4(y  - 4)^2/4=4/4}}}

{{{(x+6 )^2/4+ (y  - 4)^2/1=1}}}

{{{h=-6}}} and {{{k=4}}},
{{{a=2}}} and {{{b=1}}}

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

{{{c^2=4+1}}}
{{{c=sqrt(5)}}}

The distance from the center to either focus is the fixed value {{{c}}}. The distance from the center to a vertex is the fixed value {{{a}}}.

 so center is at ({{{-6}}},{{{4}}})

the length of the whole major axis is {{{a=2}}}

the length of the whole minor axis is {{{b=1}}}

vertices: ({{{-8}}},{{{4}}}) and ({{{-4}}},{{{4}}})


{{{ graph( 600, 600, -10,5, -5, 10,sqrt(1-(x+6 )^2/4)+4, -sqrt(1-(x+6 )^2/4)+4) }}}