Question 948467
{{{x^2/36+y^2/16=1}}}

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

CENTER: Center is at (h,k) in this case (0,0). Center is at origin.

FOCI: Focus is (f) from center {{{f^2=a^2-b^2}}}
{{{f^2=36-16=20}}} Find square root of each side.
{{{f=sqrt(20)}}}
 Foci are at ({{{-sqrt(20)}}},{{{0}}}) and ({{{sqrt(20)}}},{{{0}}}). 

VERTICES: In this case at (+ or - a,0) and a={{{sqrt(36)}}}=6
  Vertices at (-6,0) and (6,0)


ENDPOINTS OF LATERA RECTA: The latera recta are perpendicular to the major axis at the foci, and have length: {{{2b^2/a}}}. Since half is above and half is below the axis, we need half the length or {{{b^2/a}}}=16/6=8/3
For the focus at ({{{-sqrt(20)}}}),{{{0}}}), the endpoints of the latus rectum are  ({{{-sqrt(20)}}},{{{(8/3)}}}) and ({{{-sqrt(20)}}},{{{-(8/3)}}})
For the focus at ({{{sqrt(20)}}},{{{0}}}), the endpoints of the latus rectum are ({{{sqrt(20)}}},{{{(8/3)}}}) and ({{{sqrt(20)}}},{{{-(8/3)}}})

ECCENTRICITY: Eccentricity {{{epsilon}}}=f/a={{{sqrt(20)/6}}}

EQUATIONS OF DIRECTRICES: directrix is a line perpendicular to the main axis
on opposite the vertex from the focus, and same distance as the focus from the vertex. The directrix is outside the ellipse. The  is a+(a-f) from the center,
or in this case, the directrix to the right of the origin is x={{{6+(6-sqrt(20))}}}={{{12-sqrt(20)}}} and to the left of the origin x=-({{{12-sqrt(20)}}}).