How to prove that the angle $\angle AMB$ gets its maximum when M is the intersection of y-axis and the minor axis of the ellipse? 
A and B are the intersections of x-axis and major axis of the ellipse, M is a point on the ellipse.
 A: With a slight renaming of elements, consider a point $M$ on an ellipse with vertices $P$ and $Q$, major radius $a$, and minor radius $b$. Let $N$ be the foot of the perpendicular from $M$ to the major axis, and define $p := |\overline{PN}|$, $q:=|\overline{QN}|$, $m := |\overline{MN}|$. Also, $\theta = \angle MPQ$ and $\phi := \angle MQP$.

We may regard such an ellipse as the result of a circle of radius $a$ having been subjected to a "vertical" scaling transformation of factor $b/a$; conversely, we can recapture the circle by scaling the ellipse, in the direction perpendicular to its major axis, by factor $a/b$. Let $M^\prime$ be the image of $M$ on the circle, so that $m^\prime := |\overline{M^\prime N}| = m a/b$. Observe that, because $\triangle PM^\prime Q$ has a right angle at $M^\prime$, $m^\prime$ is the geometric mean of $p$ and $q$. Thus,
$$(m^\prime)^2 = p q = \frac{m}{\tan\theta} \frac{m}{\tan\phi} \qquad\to\qquad \tan\theta\tan\phi = \left(\frac{m}{m^\prime}\right)^2 = \frac{b^2}{a^2} \tag{1}$$
But, then, with $\tan= \sin/\cos$, we can write
$$\frac{a^2-b^2}{a^2+b^2} = \frac{\cos\theta\cos\phi - \sin\theta\sin\phi}{\cos\theta\cos\phi + \sin\theta\sin\phi} = \frac{\cos(\theta+\phi)}{\cos(\theta-\phi)} \tag{2}$$
Since the left-hand side of $(2)$ is constant, the elements of ratio on the right-hand must be simultaneously maximized. For $\cos(\theta-\phi)$, maximization clearly happens when $\theta = \phi$. Now, for the angle ranges in effect, maximizing a cosine is equivalent to minimizing its argument. We have, then, that $\theta+\phi$ is smallest ---and target angle $\angle PMQ$ ($=180^\circ-\theta-\phi$) is largest--- when $\triangle PMQ$ is isosceles; that is, when $M$ is an endpoint of the ellipse's minor axis. $\square$
A: The locus of points such that $\angle AMB = \theta $ is the arc of a circle passing through $A,B$
Consider the family of circles passing through $A,B$. For the 'auxiliary circle' i.e. the circle with $A,B$ as diameter, any point on this circle would form a right angle on the circumference. This circle only has $A,B$ as points on the circumference, and the centre lies on the axis of the ellipse.
Now, as the centre of the circle passing through $A,B$ lowers , there are two points of intersection with ellipse, and the angle $AMB$ increases. By this  process $AMB$ reaches a maximum when the circle through $A,B$ is tangent at the end of the minor axis, which is the desired result
A: Use cosine rule and some facts 
We can use the fact that $AM+BM=2a$ for all $M$ on the ellipse.
Then write the angle AMB in terms of AM,BM and 2a using cosine rule 
 $(2a)^2 = (AM)^2+(BM)^2-2(AM)(BM)\cos(AMB)$ 
Then calculate $\frac {d(AMB)}{dx}$ and equate it to zero if $AM=x$ and $BM=2a-x$
Edit 1 : My answer was wrong ... It would be right if $A$ and $B$ are focii of the ellipse
A: Let $N$ be the point where line $AM$ meets again the circle of diameter $AB$. We have $\angle AMB=90°+\theta$, where $\theta=\angle NBM$, so $\angle AMB$ is maximum if $\theta$ is maximum.
But $\theta=90°-\alpha-\beta$ (see diagram below) and if $(x,y)$ are the coordinates of $M$ we have:
$$
\tan\alpha={y\over a+x},\quad
\tan\beta={y\over a-x},\quad
\tan\theta={1\over\tan(\alpha+\beta)}=
{1-\tan\alpha\tan\beta\over\tan\alpha+\tan\beta}
={a^2-x^2-y^2\over 2ay},
$$
where $a=AB/2$ is the semi-major axis of the ellipse. 
But $x$ and $y$ are connected by the ellipse equation $x^2/a^2+y^2/b^2=1$, where $b$ is the semi-minor axis. Substituting $x^2$ form this equation into the previous formula finally yields:
$$
\tan\theta={a^2/b^2-1\over 2a}y.
$$
As $0°\le\theta<90°$, $\theta$ attains its maximum value when $\tan\theta$ does, that is when $y$ has its maximum value $b$.

