Eccentricity of an ellipse under certain conditions An ellipse inscribed in a semicircle touches the circular arc at two distinct points and also you touches the bounding diameter,it's major axis is parallel to the bounding diameter. When the ellipse has the maximum area.its eccentricity is------

The above picture shows the case when the area of the ellipse is maximum.It is quite easy to see that $2b=R$ where $b$ is the semi conjugate diameter of the ellipse and R is the radius of the semicircle.but I could not find out a relation between the semi major diameter and $R$ so that I could not find out the maximum area of ellipse by differentiating it with respect to $R$.
Although the question says that the ellipse touches the circular arc at just two points but we may assume that ellipse may remain very close to the circle such that portion of ellipse beneath the major axis is just the same as the circular arc above it.this is in my view the extreme case which makes area of ellipse maximum.
Note that for this to happen the major axis must be bisecting the radius perpendicularly and so $2b=R$.
Any help appreciated. Thanks.
 A: (see figure below)
This issue being independant of the size of the semi-circle, we will take its radius $= 1$.
The axes of the ellipses we are going to consider are parallel to the coordinate's axes.  Therefore, if we denote classically their semi-axes lengths by $a$ and $b$ $(b \leq a)$, knowing that their center is in $(0,b)$, one has:
$$\tag{1} \dfrac{x^2}{a^2}+\dfrac{(y-b)^2}{b^2}=1 \ \ \Longleftrightarrow \ \ b^2x^2+a^2y^2-2a^2by=0$$
How can we transcript the tangency of the ellipse (1) with the unit circle ?
By expressing that the system formed by (1) and the equation of the unit circle $x^2+y^2=1$ has a double root. 
Let us plug $x^2=1-y^2$ in (1), giving the equation in variable $y$:
$$\tag{2}b^2(1-y^2)+a^2y^2-2a^2by=0$$ which is the following quadratic:
$$\tag{3}(a^2-b^2)y^2-2a^2by+b^2=0.$$
(3) has a double root if its discriminant is $0,$ which is equivalent to:
$$b^2(a^4-a^2+b^2)=0$$
As $b=0$ is not to be considered (it would generate an ellipse reduced to the diameter of the semicircle!), we deduce that 
$$\tag{4}a^4-a^2+b^2=0 \ \iff \ (a^2-\frac12)^2=\frac14-b^2 \ \iff \ a^2=\frac{1\pm \sqrt{1-4b^2}}{2}$$
Only the $+$ case should be considered in (4).
$$\tag{5}a^2=\frac{1+ \sqrt{1-4b^2}}{2}$$
Maximizing the area $\pi a b$ of an ellipse is like maximizing $a^2b^2$.
Using (5), the maximum will occur for a value of $b$ such that $f(b):=(1+ \sqrt{1-4b^2})b^2$ is maximal.
But $f$ defined in this way is increasing (resp. decreasing) on $[0,\dfrac{\sqrt{2}}{3}]$ (resp. $[\dfrac{\sqrt{2}}{3},1/2]$) (see second figure below).
Thus the ellipse with maximal area has 

$b=\dfrac{\sqrt{2}}{3}\approx 0.4714$ and therefore (using (5)) $a=\sqrt{\frac23} \ \approx \ 0.8145$ (see blue curve on the figure).

Remark: using (1), the general equation of the inscribed ellipses is:
$$\tag{6}2b^2x^2+(1+\sqrt{1-4b^2})(y^2-2by)=0 \ \ \ \ (0 \leq b \leq 1/4)$$


