Question 1186658

An ellipse has vertices ({{{2 -sqrt(61)}}}, {{{5}}}) and ({{{2 + sqrt(61)}}}, {{{5}}}), and its minor axis is {{{12}}} units long. 

Find its standard equation and its foci.

center is half way between vertices

({{{(2 -sqrt(61)+2 + sqrt(61))/2}}},{{{( 5+5)/2}}})=({{{4/2}}},{{{10/2}}})=({{{2}}},{{{5}}})=>{{{h=2}}} and {{{k=5}}}

 
minor axis is {{{2b=12}}} =>{{{ b=6}}}


{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}........plug in known

{{{(x-2)^2/a^2+(y-5)^2/6^2=1}}}

{{{(x-2)^2/a^2+(y-5)^2/36=1}}}...............plug in coordinates of vertices ({{{2 -sqrt(61)}}}, {{{5}}})


{{{(2 -sqrt(61)-2)^2/a^2+(5-5)^2/36=1}}}

{{{( -sqrt(61))^2/a^2+(0)^2/36=1}}}

{{{61/a^2=1}}}

{{{61=a^2 }}}


so, your equation is:

{{{(x-2)^2/61+(y-5)^2/36=1}}}



for an ellipse with major axis parallel to the x-axis, the Foci (focus ) are defined as :
 ({{{h+c}}}, {{{k}}} ),  ({{{h-c}}}, {{{k}}} ) 

find {{{c}}}  

{{{c=sqrt(a^2-b^2)}}}

{{{c=sqrt(61-36)}}}

{{{c=sqrt(25)}}}

{{{c=5}}} or {{{c=-5}}}

({{{2+5}}},{{{5}}})=({{{7}}}, {{{5}}})

({{{2-5}}},{{{ 5}}}) =({{{-3}}},{{{5}}}) 



{{{ drawing( 600, 600, -10, 15, -10, 15,
circle(2,5,.12), locate(2,5,C(2,5)),
circle(-3,5,.12), locate(-3,5,F(-3,5)),
circle(7,5,.12), locate(7,5,F(7,5)),
circle(2 -sqrt(61),5,.12), locate(2 -sqrt(61),5,V(2 -sqrt(61),5)),
circle(2 +sqrt(61),5,.12), locate(2+sqrt(61),5,V(2 +sqrt(61),5)),
graph( 600, 600, -10, 15, -10, 15, (1/61)(305-6sqrt(61)*sqrt(-x^2+4x+57)), (1/61)(6sqrt(61)*sqrt(-x^2 +4x +57) + 305))) }}}