Question 874025
There is a typo somewhere.
If an ellipse is given by the equation {{{x^2/a^2+y^2/b^2=1}}} , with {{{a>0}}} and {{{b>0}}} ,
and there is at least one point whose distance from the center of the ellipse is equal to
{{{sqrt(a^2+b^2)/sqrt(2)}}} ,
the ellipse is a circle, with radius {{{a=b=sqrt(a^2+b^2)/sqrt(2)}}} ,
and all the infinite number of points on the ellipse are at a distance of
{{{sqrt(a^2+b^2)/sqrt(2)}}} from the center of that ellipse, which is a circle.
 
The points on an ellipse whose distance from the center of the ellipse are equal to a given distance are all the points of a circle,
with that distance as the radio, and centered at the center of the ellipse.
It could be that circle and ellipse share no points :
{{{drawing(400,400,-.4,.4,-.4,.4,grid(0),
red(arc(0,0,.6,.2,0,360)),green(circle(0,0,.35)) )}}} or {{{drawing(400,400,-.4,.4,-.4,.4,grid(0),
red(arc(0,0,.6,.2,0,360)),green(circle(0,0,.075)) )}}} .
It could be that circle and ellipse share exactly two points :
{{{drawing(400,400,-.4,.4,-.4,.4,grid(0),
red(arc(0,0,.6,.2,0,360)),green(circle(0,0,.3)) )}}} or {{{drawing(400,400,-.4,.4,-.4,.4,grid(0),
red(arc(0,0,.6,.2,0,360)),green(circle(0,0,.1)) )}}} .
It could be that circle and ellipse share exactly four points :
{{{drawing(400,400,-.4,.4,-.4,.4,grid(0),
red(arc(0,0,.6,.2,0,360)),green(circle(0,0,.2)) )}}} ,
or it could be that the ellipse IS the same curve as the circle, and so they share all their infinite number of points.
 
The ellipse {{{x^2/a^2 + y^2/b^2=1 }}} , with {{{a>0}}} and {{{b>0}}} , is centered on the origin,
so we are dealing with an ellipse and a circle centered (both of them) at (0,0), the origin.
 
The points the problem talks about are on the ellipse and on that circle.
If those points are exactly 2 points, they are either the vertices or the co-vertices.
That means, those points must be either (a,0) and {-a,0),
which are at a distance {{{a}}} from the origin,
or those points must be either (0,b), and (0,-b),
which are at a distance {{{b}}} from the origin.
 
If the distance is {{{ sqrt(a^2+b^2)/sqrt(2) }}} , as you posted,
then {{{a=sqrt(a^2+b^2)/sqrt(2) }}} or {{{b=sqrt(a^2+b^2)/sqrt(2) }}} .
However,
{{{a=sqrt(a^2+b^2)/sqrt(2)}}}-->{{{a^2=(a^2+b^2)/(sqrt(2)^2)}}}-->{{{a^2=(a^2+b^2)/2}}}-->{{{2a^2=a^2+b^2}}}-->{{{a^2=b^2}}}-->{{{a=b}}} ,
and
{{{b=sqrt(a^2+b^2)/sqrt(2)}}}-->{{{b^2=(a^2+b^2)/(sqrt(2)^2)}}}-->{{{b^2=(a^2+b^2)/2}}}-->{{{2b^2=a^2+b^2}}}-->{{{b^2=a^2}}}-->{{{b=a}}} .
Either way, the ellipse turns to be a circle of radius {{{a=b}}} ,
and all of its infinite number of points are at a distance
{{{sqrt(a^2+b^2)/sqrt(2)=sqrt(a^2+a^2)/sqrt(2)=sqrt(2a^2)/sqrt(2)=sqrt(2)*sqrt(a^2)/sqrt(2)=sqrt(a^2)=a=b}}} .
 
NOTE - CHECK THE PROBLEM'S EQUATIONS:
If the distance were
{{{sqrt(a^2+b^2)/2}}} , then
{{{a=sqrt(a^2+b^2)/2}}}-->{{{a^2=(a^2+b^2)/2^2}}}-->{{{a^2=(a^2+b^2)/4}}}-->{{{4a^2=a^2+b^2}}}-->{{{3a^2=b^2}}} ,
and if
{{{b=sqrt(a^2+b^2)/2}}}-->{{{b^2=(a^2+b^2)/2^2}}}-->{{{b^2=(a^2+b^2)/4}}}-->{{{4b^2=a^2+b^2}}}-->{{{3b^2=a^2}}} .
 
In either case, the eccentricity would be the same.
 
If {{{3b^2=a^2}}}<-->{{{3b^2=a^2}}}  , then {{{a>b>0}}} ;
{{{a=sqrt(3b^2)=sqrt(3)*b}}} is the semi-major axis;
the vertices are (a,0) and {-a,0);
{{{c^2=a^2+b^2=3b^2+b^2=4b^2}}} --> c+sqrt(4b^2)=2b}}} ,
and the eccentricity is {{{c/a=2b/(sqrt(3)*b)=2/sqrt(3)=2sqrt(3)/3}}} .
 
If {{{3a^2=b^2}}}<-->{{{b^2=3a^2}}}  , then {{{b>a>0}}} ;
{{{b=sqrt(3a^2)=sqrt(3)*a}}} is the semi-major axis;
the vertices are (0,b) and {0,-b);
{{{c^2=a^2+b^2=a^2+3a^2=4a^2}}} --> {{{c=sqrt(4a^2)=2a}}} ,
and the eccentricity is {{{c/b=2a/(sqrt(3)*a)=2/sqrt(3)=2sqrt(3)/3}}}