Question 1136850
Find the standard form of the equation of the ellipse with the given characteristics. 

Center: ({{{6}}}, {{{7}}}); 
{{{a = 3c}}}; 
foci: ({{{2}}}, {{{7}}}), ({{{10}}}, {{{7}}}) 


The equation of an ellipse is :

{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}} for a horizontally oriented ellipse and 

{{{(x-h)^2/b^2+(y-k)^2/a^2=1}}} for a vertically oriented ellipse.

({{{h}}},{{{k}}}) is the center and the distance{{{ c}}} from the center to the foci is given by {{{a^2-b^2=c^2}}}
 
{{{a}}} is the distance from the center to the vertices and {{{b }}}is the distance from the center to the co-vertices.

The center of the ellipse is ({{{6}}}, {{{7}}})=>{{{h=6}}} and {{{k=7}}}

{{{a}}} is the distance between the center and the vertices, so {{{a=9}}}

{{{c}}} is the distance between the center and the foci, and foci is at ({{{2}}}, {{{7}}}), ({{{10}}}, {{{7}}})

distance from {{{6}}} to {{{2}}} is {{{4}}}, and distance from {{{6}}} to {{{10}}} is {{{4}}}
so {{{c=4}}}

since {{{a = 3c}}}, we have {{{a = 3*4}}}=>{{{a = 12}}}

now find {{{b}}}

{{{12^2-b^2=4^2}}}
{{{12^2-4^2=b^2}}}
{{{144-16=b^2}}}
{{{b^2=128}}}

{{{(x-6)^2/144+(y-7)^2/128=1}}} for a horizontally oriented ellipse


{{{drawing( 600, 600, -10, 25, -10, 25,
circle(6,7,.12),circle(2,7,.12),circle(10,7,.12),
locate(6,7,C(6,7)),locate(2,7,F(2,7)),locate(10,7,F(10,7)),
 graph( 600, 600, -10, 25, -10, 25,- sqrt((1-(x-6)^2/144)128)+7 ,sqrt((1-(x-6)^2/144)128)+7) )}}}