As the center is the midpoint of the foci, so the center $O(0,0)$
As the foci lie on the major axis , so the equation of the major axis $y=0\implies$ the equation of the minor axis $x=0$
Now, if the length of the major, minor axes be $2a,2b$ respectively with eccentricity $=e$
So, the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Any point $P$ on the ellipse can be $P(a\cos\theta,b\sin\theta)$
So, the distance between $(a\cos\theta,b\sin\theta), (ae,0)$ is
$\sqrt{(ae-a\cos\theta)^2+(b\sin\theta-0)^2}$
$=\sqrt{a^2e^2+a^2\cos^2\theta-2a^2e\cos\theta+a^2(1-e^2)(1-\cos^2\theta)}$
$=a(1-e\cos\theta)$ as $0\le e<1,-1\le \cos\theta\le 1$ and $b^2=a^2(1-e^2)$
Similarly, the distance between $(a\cos\theta,b\sin\theta), (-ae,0)$ is $=a(1+e\cos\theta)$
As you know, the foci of an ellipse whose equation is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$are described, if $a^2>b^2$, by the coordinates $(c,0)$ and $(-c,0)$ where $c=\sqrt{a^2-b^2}$. In fact the sum of the distances of a generical point $(x,y)$ from $(c,0)$ and $(-c,0)$ is, as we can see by using the Pythagorean theorem, $$\sqrt{(x-c)^2+(y-0)^2}+\sqrt{(x-(-c))^2+(y-0)^2}$$which we can prove to be the constant $2a$ (assuming $a>0$). In fact, if we plug $c=\sqrt{a^2-b^2}$ into the equation of the ellipse, it becomes $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$$which is equivalent to $$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$$which is in turn equivalent, as we can see by adding $-2a^2cx$ to both members and rearranging the addends, to $$a^2((x-c)^2+y^2)=a^2(x^2-2cx+c^2+y^2)=a^4-2a^2cx+c^2x^2=(a^2-cx)^2$$which becomes, if we calculate the square root of both members and multiplicate by $4$ $$\pm 4a\sqrt{(x-c)^2+y^2}=4(a^2-cx)$$ but, since $c=\sqrt{a^2-b^2}<a$ and $\frac{x^2}{a^2}\le 1$ (see equation of the ellipse: $\frac{y^2}{b^2}\ge 0$) and therefore, for $x>0$, $cx\le ax\le a^2$, we can chose the sign + in front of the square root:$$0=4(a^2-cx)-4a\sqrt{(x-c)^2+y^2}$$which is in turn equivalent, as we notice if we add $x^2+2cx+c^2+y^2$ to both members, to $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2-4a\sqrt{(x-c)^2+y^2}=\left(2a-\sqrt{(x-c)^2+y^2}\right)^2$$which is finally equivalent, in turn, as we see by calculating the square root of both memebrs, to$$\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2}$$where we chose the positive sign for the square root $\sqrt{(x+c)^2+y^2}$ because $2a-\sqrt{(x-c)^2+y^2}\ge 0$, in fact if $2a<\sqrt{(x-c)^2+y^2}$ then $4a^2<(x-c)^2+y^2$ and $3a^2<x^2+y^2-cx-b^2$ which is impossible because if $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $a^2>b^2$ then $\frac{x^2+y^3}{a^2}<1$. The last equality means that the sum of the distances of any point from $(c,0)$ and $(-c,0)$ is $2a$, as we wanted to show.
If $b^2>a^2$ the roles of $x$ and $y$ are inverted and therefore the foci are $(0,\sqrt{b^2-a^2})$ and $(0,-\sqrt{b^2-a^2})$.
Best Answer
HINT
J = center of ellipse
Radius of circle = a
E1 '= second focus