Question 1120509
 Find an equation for the ellipse with:

 center({{{5}}}, {{{0}}}), 
foci({{{5}}}, ±{{{4}}})  and 
major axis of length {{{10}}}

Because the {{{y}}} coordinate of the foci is the coordinate that is changing, we know that the major axis of the ellipse is parallel to the {{{y}}} axis. Therefore, the standard Cartesian form of the equation of the ellipse is:


{{{(y-k)^2 /a^2 + (x-h)^2 /b^2 = 1}}} (for a taller-than-wide ellipse )


if center is at ({{{h}}},{{{k}}})=({{{5}}},{{{0}}}) we have {{{h=5}}} and {{{k=0}}}
if major axis of length {{{10}}}, and "{{{a}}}" is half of the length of the major axis, we have  =>{{{a=5 }}}

so far we have

{{{(y-0)^2 /5^2 + (x-5)^2 /b^2 = 1 }}}


foci({{{5}}}, ±{{{4}}})

 since the focus is {{{4}}} units above the center, then {{{c = 4}}}

use the equation {{{b^2 = a^2 - c^2 }}} to find {{{b}}}

{{{b^2 = 5^2 - 4^2 }}}

{{{b^2 = 25 - 16}}}
 
{{{b^2 = 9}}}

{{{b =3}}}

and we have

 {{{y^2 /5^2+(x-5)^2 /3^2  = 1}}}


 {{{y^2 /25+(x-5)^2 /9  = 1}}}


{{{ graph( 600, 600, -10, 10, -10, 10,-sqrt(25(1-(x-5)^2 /9 )),sqrt(25(1-(x-5)^2 /9 ))) }}}