Question 1181121
Find the equation of the ellipse 

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

with center at ({{{0}}}, {{{0}}}), =>{{{h=0}}},{{{k=0}}}

the equation of the ellipse is

{{{x^2/a^2+y^2/b^2=1}}}


length of major axis is {{{2a=8}}}=>{{{a=4}}} 

and a focus at ({{{0}}}, {{{-2}}})

since a focus is at ({{{0}}}, {{{c}}})=>{{{c=2}}}
 and 
other focus is on oposite side of the center at ({{{0}}}, {{{2}}})

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

{{{b^2=4^2-(-2)^2}}}

{{{b^2=16-4}}}

{{{b^2=12}}}

{{{b=sqrt(12)}}}

{{{b=sqrt(4*3)}}}

{{{b=2sqrt(3)}}}


your equation is: major axis is on y-axis, so {{{a}}} goes with {{{y}}}

{{{(x-h)^2/b^2+(y-k)^2/a^2=1}}}
{{{x^2/12+y^2/16=1}}}


1. Center ({{{0}}}, {{{0}}})
2. Foci
𝐹1 ({{{0}}}, {{{-2}}})
𝐹2({{{0}}}, {{{2}}})

3. Vertices V =>({{{0}}},{{{a}}}),({{{0}}},{{{-a}}})
𝑉1 ({{{0}}},{{{-4}}})
𝑉2({{{0}}},{{{4}}})

4. Co-vertices =>({{{b}}},{{{0}}}),({{{-b}}},{{{0}}})
𝐵1({{{2sqrt(3)}}}, {{{0}}})
𝐵2({{{-2sqrt(3)}}}, {{{0}}})

5. Endpoints of Latus
Rectum
𝐸1 ({{{h-b^2/a}}}, {{{k+c}}})=> ({{{0-12/4}}}, {{{0+2}}})=> ({{{-3}}}, {{{2}}})
𝐸2({{{h+b^2/a}}}, {{{k+c}}})=> ({{{0+12/4}}}, {{{0+2}}})=> ({{{3}}}, {{{2}}})
𝐸3({{{h-b^2/a}}}, {{{k-c}}})=> ({{{0-12/4}}}, {{{0-2}}})=> ({{{-3}}}, {{{-2}}})
𝐸4({{{h+b^2/a}}}, {{{k-c}}})=> ({{{0+12/4}}}, {{{0-2}}})=> ({{{3}}}, {{{-2}}})

6. Directrices: {{{y=-8}}}, {{{y=8}}}

7. Eccentricity:{{{1/2}}}

8. Length of LR:{{{6}}}
First latus rectum: {{{y=-2}}}
Second latus rectum:{{{y=2}}}
The length of the latera recta: {{{6}}}

9. Length of Major Axis:{{{8}}}
10.Length of Minor Axis:{{{4}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(0, -2,.15),locate(0.3, -2,F),
circle(0, 2,.15),locate(0.3, 2,F),
circle(0, 4,.15),locate(0.3, 4,V),
circle(0, -4,.15),locate(0.3, -4,V),
circle(2sqrt(3), 0,.15),locate(2sqrt(3), 0.3,cV),
circle(-2sqrt(3), 0,.15),locate(-2sqrt(3), 0.3,cV),
circle(-3, 2,.15),locate(-3, 2,E1),
circle(3, 2,.15),locate(3, 2,E2),
circle(-3, -2,.15),locate(-3, -2,E3),
circle(3, -2,.15),locate(3, -2,E4),
graph( 600, 600, -10, 10, -10, 10,-sqrt(16(1-x^2/12)),sqrt(16(1-x^2/12)) )) }}}