Question 1168621
 find the equation of the ellipse in general form with vertices at ({{{-2}}},{{{1}}}) and ({{{4}}},{{{1}}}) with eccentricity of   {{{2/3}}}

the equation of the ellipse in general form:

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

The distance from the center to a vertex is the fixed value {{{a}}}.

{{{e = c/a}}} => {{{c/a=2/3}}}-> {{{c=2}}} and {{{a=3}}}

then 

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

{{{b ^2= 3^2 - 2^2}}}

{{{b^2= 9 - 4}}}

{{{b^2= 5}}}

{{{b= sqrt(5)}}}


so far equation is:


{{{(x-h)^2/9+(y-k)^2/5=1}}}

The distance from the center to a vertex is the fixed value {{{a}}}.

Since the vertices of the ellipse are ({{{-2}}},{{{1}}}) and ({{{4}}},{{{1}}}) , and {{{a=3}}} then

({{{-2+3}}},{{{1}}})=({{{1}}},{{{1}}})
 
({{{4-3}}},{{{1}}})=({{{1}}},{{{1}}})

=> {{{h = 1}}}, {{{k = 1}}}

and your equation is:

{{{(x-1)^2/9+(y-1)^2/5=1}}}

 

{{{drawing(600, 600, -10, 10, -10, 10,
circle(-2,1,.12),circle(4,1,.12),locate(-2,1,v(-2,1)),locate(4,1,v(4,1)),
 graph( 600, 600, -10, 10, -10, 10,-sqrt(5-(5(x-1)^2)/9) +1, sqrt(5-(5(x-1)^2)/9) +1)) }}}