Question 1091493

{{{x^2/16 - x/4 + y^2/9 + 4y/9 = 11/16}}} ..........complete squares


{{{(1/16)(x^2 - 4x)+(1/9)( y^2 + 4y) = 11/16}}} 


{{{(1/16)(x^2 - 4x+b^2)-(1/16)b^2+(1/9)( y^2 + 4y+b^2)-(1/9)b^2 = 11/16 }}}


{{{(1/16)(x^2 - 4x+2^2)-(1/16)2^2+(1/9)( y^2 + 4y+2^2)-(1/9)2^2 = 11/16 }}}


{{{(1/16)(x - 2)^2-(1/16)4+(1/9)( y + 2)^2-(1/9)4 = 11/16 }}}


{{{(1/16)(x - 2)^2-(1/4)+(1/9)( y + 2)^2-(4/9)= 11/16 }}}


{{{(1/16)(x - 2)^2+(1/9)( y + 2)^2 = 11/16 +(1/4)+(4/9)}}}


{{{(1/16)(x - 2)^2+(1/9)( y + 2)^2 = 199/144}}}


{{{((1/16)/(199/144))(x - 2)^2+((1/9)/(199/144))( y + 2)^2 = (199/144)/(199/144)}}}

{{{(9/199)(x - 2)^2+(16/199)( y + 2)^2 =1}}}


{{{(x - 2)^2/(199/9)+( y + 2)^2/(199/16) =1}}}


->{{{h=2}}} and {{{k=-2}}}, so the center is at ({{{2}}}, {{{-2}}})

semi-major axis length: 

{{{a=sqrt(199/9) =sqrt(199)/3}}}≈ {{{a=4.7}}}
semi-minor axis length:{{{ b=sqrt(199/16) =sqrt(199)/4}}}≈{{{b=3.5}}}

{{{c^2=a^2-b^2}}}
{{{c^2=199/9-199/16}}}
{{{c^2=1393/144}}}
{{{c=sqrt(1393/144)}}}
{{{c=sqrt(1393)/12}}}

foci: ({{{h - c}}}, {{{k}}}) =({{{2 - sqrt(1393)/12}}}, {{{-2}}}) ≈ ({{{-1.1}}}, {{{-2}}}) 
and
foci :({{{h +c}}}, {{{k}}}) =({{{2 +sqrt(1393)/12}}},{{{ -2}}}) ≈ ({{{5.1}}},{{{ -2}}})

vertices:

The vertices are {{{a = sqrt(199)/3}}} units above and below the center, at 
({{{2 - sqrt(199)/3}}}, {{{-2}}}) ≈ ({{{-2.7}}}, {{{-2}}})
and
({{{2 + sqrt(199)/3}}}, {{{-2}}}) ≈ ({{{6.7}}}, {{{-2}}})

The co-vertices are {{{b = sqrt(199)/4}}} units to either side of the center, at 
({{{2+sqrt(199)/4}}} , {{{-2}}}) and ({{{2-sqrt(199)/4}}}, {{{-2}}})