Question 936512
The standard form of the equation of an ellipse is:
{{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}} for horizontal ellipses, and
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}} for vertical ellipses 

In these equations, the "{{{h}}}" and the "{{{k}}}" are the {{{x}}} and {{{y}}} coordinates, respectively, of the center. 

here is your ellipse:

{{{x^2/9 + y^2/16 =1}}}

from given we know that {{{h=0}}}, {{{k=0}}}; so, center is at origin ({{{0}}},{{{0}}})

we also know that major axis is vertical and 

semi-major axis length is {{{a=sqrt(16)=4}}}

semi-minor axis length is {{{b=sqrt(9)=3}}}

the vertices will be "{{{a}}}" distance above and below the center:

vertices  ({{{0}}}, {{{-4}}})  |  ({{{0}}},{{{ 4}}})

 The co-vertices are right and left: 

({{{-3}}},{{{ 0}}})  |  ({{{3}}},{{{0}}})

The distance from the center to each focus is called "{{{c}}}". The "{{{c}}}" is not in the standard form of the equation for an ellipse. But there is a fixed relationship between the "{{{a}}}", the "{{{b}}}" and the "{{{c}}}" values:
{{{a^2=b^2+c^2}}}

{{{16=9+c^2}}}

{{{c^2=16-9}}}

{{{c^2=7}}}

{{{c=sqrt(7)}}}

foci: | ({{{0}}}, {{{-sqrt(7)}}})  |  ({{{0}}}, {{{sqrt(7)}}})

or approximately ({{{0}}}, {{{-2.65}}})  |  ({{{0}}}, {{{2.65}}})


{{{ graph( 600, 600, -5, 5, -5, 5,sqrt((1-x^2/9 )16) ,-sqrt((1-x^2/9 )16)) }}}