Question 1045317
You also need to know the coordinates of the center, and
what direction the major axis follows.
 
An ellipse is a stretched circle.
A circle of radius {{{R}}} with center at the origin has the equation
{{{x^2+y^2=R^2}}} <---> {{{x^2/R^2+y^2/R^2=1}}} ,
because {{{x^2+y^2}}} is the square of the distance from point (x,y) to origin (0,0) .
Thankfully, we almost always get confronted by ellipses that are easier to calculate,
where the longest axis of symmetry, the major axis, is either
parallel to the x-axis (and we call it horizontal), or
parallel to the y-axis (and we call it vertical).
An ellipse with center at the origin and axes of symmetry parallel to the x- and y-axes has the equation
{{{x^2/a^2+y^2/b^2=1}}} or {{{x^2/b^2+y^2/a^2=1}}} , with {{{a>b}}} .
If the center of a circle, or ellipse is not {{{O(0,0)}}},
but a point {{{C(h,k)}}} ,
you just write {{{(x-k)^2}}} instead of {{{x^2}}} ,and
{{{(y-k)^2}}} instead of {{{y^2}}} .
  
The distance {{{c}}} from each focus to the center, is related to {{{a}}} and {{{b}}} by {{{a^2=b^2+c^2}}} , so you only need to know two of those distances.
Given foci and length of major axis, you have {{{a}}} and {{{c}}} ,
just find {{{b^2=a^2-c^2}}} and plug the found {{{b^2}}} value into the equation.
 
{{{drawing(200,200,-4,4,-4,4,grid(0),circle(0,0,3),circle(0,0,0.1),
locate(0.5,-3,circle))}}} {{{drawing(300,200,-6,6,-4,4,grid(0),arc(0,0,10,6,0,360),circle(0,0,0.1),
locate(-5.5,-2.5,ellipse),locate(-5.5,-3,horizontal),
locate(-2,-3,major),locate(0.2,-3,axis))}}} {{{drawing(200,300,-4,4,-6,6,grid(0),arc(0,0,6,10,0,360),circle(0,0,0.1),
locate(-3.9,6,ellipse),locate(-3.9,5.5,vertical),
locate(-3.9,5,major),locate(-3.9,4.5,axis))}}} .
 
The points the farthest from the center are the vertices,
located on the major axis, at a distance {{{a}}} from the center.
The points the closest to the center are often called co-vertices,
located on the minor axis, at a distance {{{b}}} from the center.
The foci are located on the major axis, at a distance {{{c}}} to either side of the center.
 
Here is an ellipse. you see the origin, point {{{O(0,0)}}} .
You see one labeled vertex, point {{{A(a,0)}} , at a distance {{{a}}} from the center.
I labeled the foci, focus {{{F(c,0)}}} , and Focus {{{E(-c,0)}}} ,
both at a distance {{{c}}} from the center.
I also labeled one of the co-vertices, point {{{B(0,b)}}} , at a distance {{{b}}} from the center.
The fancy definition of ellipse says that for all the points on the ellipse,
the sum of the distances to one Focus and the other is the same. 
{{{drawing(400,400,-6,6,-4,4,grid(0),arc(0,0,10,6,0,360),circle(0,0,0.1),
circle(5,0,0.1),circle(4,0,0.1),circle(-4,0,0.1),circle(0,3,0.1),
red(triangle(0,0,0,3,4,0)),red(triangle(0,0,0,3,-4,0)),
locate(0.1,1.7,red(b)),locate(1.4,0.3,red(c)),locate(2,1.4,red(a)),
locate(5.1,0.3,A),locate(0.2,3.3,B(0,b)),locate(3.2,-0.3,F(c,0)),
locate(-4.5,-0.3,E(-c,0)),locate(0.2,-0.1,O)
)}}}
You can see that point {{{a}}} is at a distance {{{a-c}}} from focus {{{F}}} ,
and at a distance ({{{a+c}}} from focus {{{E}}}.
So the sum of the distance to the foci is
{{{a=c+a+c=2a}}} for point {{{A}}}, and for all points in the ellipse.
Then, point {{{B}}} is at the same distance {{{a}}} from focus {{{E}}} and focus {{{F}}} .
Applying the Pythagorean theorem to right triangle {{{FBO}}} ,
you get the relationship {{{a^2=b^2+c^2}}} .