Question 1136776
an equation for the ellipse with vertices at ({{{-3}}}, {{{4}}}) and ({{{15}}}, {{{4}}}); focus at ({{{13}}}, {{{4}}}) 


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 half way between the vertices. 

 so, ({{{(-3+15)/2}}}, {{{(4+4)/2}}}) => ({{{6}}},{{{ 4}}})

Thus, the center ({{{h}}},{{{k}}}) of the ellipse is ({{{6}}},{{{4}}}) and the ellipse is vertically oriented.

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

{{{c}}} is the distance between the center and the foci, so {{{c=7}}}

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

{{{9^2-b^2=7^2}}}

{{{81-49=b^2}}}

{{{b^2=32}}}

The equation is:

{{{(x-6)^2/81+(y-4)^2/32=1}}}



{{{drawing( 600, 600, -10, 20, -10, 10,
circle(-3,4,.12),circle(15,4,.12),circle(13,4,.12),circle(6,4,.12),
locate(-3,4,v(-3,4)),locate(15,4,v(15,4)),locate(13,4,f(13,4)),locate(6,4,c(6,4)),
 graph( 600, 600, -10, 20, -10, 10,- sqrt((1-(x-6)^2/81)32)+4 ,sqrt((1-(x-6)^2/81)32)+4) )}}}