SOLUTION: Q.there are exactly 2 points on the ellipse {{{ x^2/a^2 + y^2/b^2=1 }}} whose distance from the center of the ellipse are equal to {{{ sqrt(a^2+b^2)/sqrt(2) }}} find the eccentrici

Question 874025: Q.there are exactly 2 points on the ellipse whose distance from the center of the ellipse are equal to find the eccentricity of the ellipse.
here's what i tried:
Ans)since the ellipse is symmetric about x and y axis if there are only 2 points which satisfy a condition, they must lie on the x and y axis. which means that the two points are extremities of the ellipse on y-axis. thus on solving given distance = 2b we get e=sqrt(6/7) which is not the right answer

Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
There is a typo somewhere.
If an ellipse is given by the equation , with and ,
and there is at least one point whose distance from the center of the ellipse is equal to
,
the ellipse is a circle, with radius ,
and all the infinite number of points on the ellipse are at a distance of
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 :
or .
It could be that circle and ellipse share exactly two points :
or .
It could be that circle and ellipse share exactly four points :
,
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 , with and , 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 from the origin,
or those points must be either (0,b), and (0,-b),
which are at a distance from the origin.

If the distance is , as you posted,
then or .
However,
-->-->-->-->--> ,
and
-->-->-->-->--> .
Either way, the ellipse turns to be a circle of radius ,
and all of its infinite number of points are at a distance
.

NOTE - CHECK THE PROBLEM'S EQUATIONS:
If the distance were
, then
-->-->-->--> ,
and if
-->-->-->--> .

In either case, the eccentricity would be the same.

If <--> , then ;
is the semi-major axis;
the vertices are (a,0) and {-a,0);
--> c+sqrt(4b^2)=2b}}} ,
and the eccentricity is .

If <--> , then ;
is the semi-major axis;
the vertices are (0,b) and {0,-b);
--> ,
and the eccentricity is
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