Question 957784
<pre>{{{drawing(400,320,-5,5,-4,4,
 
line(-7,0,7,0), line(0,-7,0,7), arc(0,0,8,-4sqrt(2)),
green(line(2sqrt(2),2,2sqrt(2),-2)),
red(circle(2.82842713,0,0.15),circle(2.82842713,0,0.13),circle(2.82842713,0,0.11),circle(2.82842713,0,0.09),circle(2.82842713,0,0.07),circle(2.82842713,0,0.05),circle(2.82842713,0,0.03),circle(2.82842713,0,0.01)),
locate(2.8,2.3,(matrix(1,3,2sqrt(2),",",2))),
red(circle(-2.82842713,0,0.15),circle(-2.82842713,0,0.13),circle(-2.82842713,0,0.11),circle(-2.82842713,0,0.09),circle(-2.82842713,0,0.07),circle(-2.82842713,0,0.05),circle(-2.82842713,0,0.03),circle(-2.82842713,0,0.01))
 
 
 

 )}}}
 
In the graph above, the green line is the latus rectum, the two red
points are the foci.

The distance between the foci is given as {{{4sqrt(2)}}}
 
Since c is half the distance between foci, {{{c = 2sqrt(2)}}}.

Since the latus rectum is given as 4, the y-coordinate of the
point at the top of the latus rectum is 2. So the coordinates
of the point at the top of the latus rectum are {{{(matrix(1,3,2sqrt(2),",",2))}}}.

The equation of the ellipse is

(1)   {{{x^2/a^2+y^2/b^2}}}{{{""=""}}}{{{1}}}

Substituting that point:

{{{(2sqrt(2))^2/a^2+(2)^2/b^2}}}{{{""=""}}}{{{1}}}

{{{8/a^2+4/b^2}}}{{{""=""}}}{{{1}}}

(2)  {{{8b^2+4a^2}}}{{{""=""}}}{{{a^2b^2}}}

Since {{{c^2=a^2-b^2}}} and {{{c = 2sqrt(2)}}}

{{{(2sqrt(2))^2=a^2-b^2}}}
{{{8=a^2-b^2}}}
{{{b^2=a^2-8}}}

Substituting in (2)

{{{8(a^2-8)+4a^2}}}{{{""=""}}}{{{a^2(a^2-8)}}}

{{{8a^2-64+4a^2}}}{{{""=""}}}{{{a^4-8a^2}}}

{{{12a^2-64}}}{{{""=""}}}{{{a^4-8a^2}}}

{{{-64}}}{{{""=""}}}{{{a^4-20a^2}}}

{{{"0"}}}{{{""=""}}}{{{a^4-20a^2+64}}}

{{{"0"}}}{{{""=""}}}{{{(a^2-16)(a^2-4)}}}

{{{"0"}}}{{{""=""}}}{{{(a-4)(a+4)(a-2)(a+2)}}}

That has solutions a=4, a=-4, a=2, a=-2

The only one it can be is a=4

Substituting in {{{b^2=a^2-8}}}

{{{b^2=4^2-8}}}
{{{b^2=16-8}}}
{{{b^2=8}}}

So the equation of the ellipse, substituting in (1),

{{{x^2/a^2+y^2/b^2}}}{{{""=""}}}{{{1}}}

{{{x^2/4^2+y^2/8}}}{{{""=""}}}{{{1}}}

{{{x^2/16+y^2/8}}}{{{""=""}}}{{{1}}}

Edwin</pre>