Question 1112231
For a wider-than-tall ellipse with center at ({{{h}}},{{{ k}}}), having vertices {{{a}}} units to either side of the center and foci {{{c}}} units to either side of the center, the ellipse equation is:
 
{{{(x-h)^2/a^2 + (y-k)^2/b^2=1}}}

For a taller-than-wide ellipse with center at ({{{h}}},{{{ k}}}), having vertices {{{a}}} units above and below the center and foci {{{c }}}units above and below the center, the ellipse equation is:

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


The length of the semi-major axis is {{{a}}}, a "semi-minor" axis is {{{b}}},  the length of the whole major axis is {{{2a}}},the length of the whole minor axis is {{{2b}}}, and the distance between the foci is {{{2c}}}. 

The three letters {{{a}}}, {{{b}}}, and {{{b}}} are related by the equation {{{b^2 = a^2 - c^2}}} .

you are given:

{{{x^2/4 + y^2/16=1}}}  or {{{ y^2/16 +x^2/4 =1}}}


you also see that  {{{a^2=16}}}-> {{{a=4}}} and {{{b^2=4}}}-> {{{b=2}}} 

as you can see, {{{h=0}}} and {{{k=0}}}; so, the center of your ellipse is at origin  ({{{0}}}{{{0}}})


now we can find the foci, eccentricity, length of latus rectum, and the x and y intercepts:

center is at:  ({{{h}}}, {{{k}}})
({{{0}}}, {{{0}}})

  focus is the fixed value {{{c}}}
({{{0}}},{{{c}}}) and ({{{0}}},{{{-c}}})

use  {{{b^2 = a^2 - c^2}}} to find {{{c}}}
{{{ c^2=a^2 -b^2 }}}
{{{ c^2=16 -4}}}
{{{ c^2= 12 }}}
{{{ c^2=4*3 }}}
{{{ c= sqrt(4*3) }}}
{{{ c= 2sqrt(3)) }}} or {{{ c= -2sqrt(3)) }}}

so, foci is at:

({{{0}}}, {{{2sqrt(3)}}}) and ({{{0}}}, {{{-2sqrt(3)}}})

vertices at: ({{{0}}}, {{{-a}}}) and  ({{{0}}}, {{{a}}})
 ({{{0}}}, {{{-4}}}) and  ({{{0}}}, {{{4}}})

covertices at:  ({{{b}}}, {{{0}}})  and ({{{-b}}}, {{{0}}})
 ({{{2}}}, {{{0}}})  and ({{{-2}}}, {{{0}}})

eccentricity is denoted as {{{e = c/a}}} 
which is
 {{{e= 2sqrt(3)/4}}}
{{{e= sqrt(3)/2}}}
 ≈ {{{0.866025}}}

length of latus rectum:  {{{LR=2b^2/a}}}=> {{{LR=(2*4)/4=2}}}

{{{x}}}-intercept: set {{{y=0}}}

{{{x^2/4 + 0^2/16=1}}}
{{{x^2/4 =1}}}
{{{x^2 =4}}}
{{{x =2}}} or {{{x =-2}}}

{{{x}}}-intercepts are at ({{{2}}}, {{{0}}}) and ({{{-2}}}, {{{0}}})

{{{y}}}-intercept: set {{{x=0}}}

{{{0^2/4 + y^2/16=1}}}
{{{y^2/16 =1}}}
{{{y^2 =10}}}
{{{y =4}}} or {{{y =-4}}}

{{{x}}}-intercepts are at ({{{0}}}, {{{4}}}) and ({{{0}}}, {{{-4}}})


{{{ graph( 600, 600, -10, 10, -10, 10,sqrt(16-4x^2) , -sqrt(16-4x^2)) }}}