Question 857851
If you know basic algebra and learn the meaning of "ellipse" and related words, such as "vertices", "foci", and "major axis", you know all you need to know about ellipses.

Looking at {{{x^2/16 + y^2/64 = 1}}} ,
you notice the following:
 
1) The ellipse is symmetrical with respect to the x- and y-axes (which means its center is (0,0), the origin).
You know that because if the coordinates of a point (p,q) satisfy the equation,
so will the coordinates of the symmetrical points (p,-q) and (-p,q).
 
2) When {{{x=0}}}, the first term is zero, and you get the most extreme values of {{{y}}} , {{{system(y=8, "or", y=-8)}}} .
Those are the coordinates of the points on the ellipse that are farthest from the center, (0,8) and (0,-8).
They are the ends of the major axis; they are called vertices,
and their distance to the center is the semi-major axis,
{{{a=8}}} .
 
3) When {{{y=0}}}, the first term is zero, and you get the most extreme values of {{{x}}} ,{{{system(x=4, "or", x=-4)}}} .
They are the ends of the minor axis, sometimes called co-vertices,
and their distance to the center is the semi-minor axis, {{{b=4}}} .
 
The foci are on the major axis (with {{{x=0}}} , as the center and vertices),
and their distance to the center, {{{c}}} , can be calculated from the formula
{{{a^2=b^2+c^2}}} (NOTE below tells you why).
In this case,
{{{8^2=4^2+c^2}}}-->{{{64=16+c^2}}}-->{{{64-16=c^2}}}-->{{{48=c^2}}}
So {{{c=sqrt(48)}}}-->{{{c=4sqrt(3)}}}-->{{{c=about6.93}}}.
That means that for the foci, {{{system(y=4sqrt(3), "or", y=-4sqrt(3))}}}.
The foci are at {{{"(0," -4sqrt(3)}}}{{{")"}}} and {{{"(0,"}}}{{{4sqrt(3)}}}{{{")"}}} .
 
NOTE:
{{{a^2=b^2+c^2}}} is not new to you.
It is the Pythagorean relationship between the sides of a right triangle.
If you look at the definition of ellipse:
"the locus of the points on a plane such that
the sum of their distances to two fixed points called foci,
equals a constant,"
and at the distances in the figure below,
you find the right triangle involved.
You find that the sum of distances for one of the vertices is {{{2a}}},
so for the co-vertices, the distance to each focus should be {{{a}}} .
So {{{a}}} is the hypotenuse of a right triangle with legs {{{b}}} and {{{c}}} .
{{{drawing(300,300,-10,10,-10,10,
grid(0),red(arc(0,0,8,16,0,360)),
blue(circle(0,0,0.2)),locate(0.1,3.7,blue(c)),
blue(circle(0,8,0.2)),blue(circle(0,-8,0.2)),
blue(circle(0,6.93,0.2)),blue(circle(0,-6.93,0.2)),
blue(circle(4,0,0.2)),blue(circle(-4,0,0.2)),
blue(triangle(0,0,0,-6.93,4,0)),blue(triangle(0,6.93,0,0,4,0)),
blue(rectangle(0,0,0.5,0.5)),locate(2.1,0.9,blue(b)),
blue(arrow(-9.8,8,-0.2,8)),locate(-9.8,8.9,blue(vertex)),
green(arrow(-6.5,8,-6.5,0)),green(arrow(-6.5,0,-6.5,8)),
green(arrow(-0.2,-8,-12,-8)),locate(-6.6,4.5,a),locate(-6.6,-3.5,a),
green(arrow(-6.5,-8,-6.5,0)),green(arrow(-6.5,0,-6.5,-8)),
locate(2.1,4.3,blue(a)),locate(0.1,-2.5,blue(c)),
blue(arrow(6,6.93,0.2,6.93)),locate(6.1,7.4,blue(focus))
)}}}