Question 1138080


Find the center, vertices, and foci of the ellipse with equation

{{{ x^2/144 + y^2/225 = 1}}}

compare to standard formula: {{{ (x-h)^2/a^2 + (y-k)^2/b^2 = 1}}}

so, your ellipse is centered at origin; {{{h=0}}},{{{k=0}}}

also, 
{{{a^2=144}}}=>{{{a=12}}}
{{{b^2=225}}}=>{{{b=15}}}

{{{c }}}= distance to focus 

For an ellipse with major axis parallel to the y-axis, the Foci (focus points ) are defined as : 
({{{h}}},{{{ k+c}}} ),  ({{{h}}}, {{{k-c}}} ),  where {{{c= sqrt(b^2-a^2)}}} is the distance from the center : ({{{h}}}, {{{k }}})  to a focus 
 
{{{c= sqrt(b^2-a^2)}}}

 {{{c= sqrt(225-144)}}}

{{{c=sqrt(81)}}}


{{{c=9}}}

so, a focus will be at
({{{0}}},{{{ 0+9}}} ),  ({{{0}}}, {{{0-9}}} )

({{{0}}},{{{ 9}}} ),  ({{{0}}}, {{{-9}}} )

the vertices are at ({{{h}}},{{{ k+b}}} ),  ({{{h}}}, {{{k-b}}} )

we have {{{b=15}}}

the vertices are at ({{{0}}},{{{ 15}}} ),  ({{{0}}}, {{{-15}}} )


answer: Center: ({{{0}}}, {{{0}}}); Vertices: ({{{0}}}, {{{-15}}}), ({{{0}}}, {{{15}}}); Foci: ({{{0}}},{{{ -9}}}), ({{{0}}}, {{{9}}})