The maximum from a point outside an ellipse to a ellipse. In the $xOy$ axes, Assume there is an ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, and a point $A(0,t)$ ($t$ is a constant )outside the ellipse. Assume $P$ is a point in the ellipse. Find the maximum of $|PA|$. and the point when obtain the maximum. 

My approach:
I consider the parameter function of ellipse: 
\begin{matrix}
x = a\cos \theta \\ 
y = b\sin \theta
\end{matrix}
then we get : $|PA|^2 = (t-b\sin \theta)^2 + a^2\cos^2 \theta$. apparently, $|PA| = a^2 + t^2+(b^2-a^2)\sin ^2 \theta -2bt \sin \theta = f(\theta)$ 
$$f'(\theta) = 2(b^2-a^2)\sin \theta \cos \theta - 2bt \cos \theta $$ then $\cos \theta = 0$ or $\sin \theta = \frac{bt}{b^2-a^2}$.
If we draw a picture, the apparently solution is $\theta = \frac{3\pi}{2}$,this is from $\cos \theta = 0$. and I think this is the only solution. But if $\left|\frac{bt}{b^2-a^2}\right|\leq 1$. Does this mean there is another solution $\theta = \arcsin\left(\frac{bt}{b^2-a^2}\right)$ ?

Is there an geometry way to see this conclusion : the maximum is obtain at $(0,-b)$? thanks very much
 A: If the bottom point of the ellipse, $(0,-b)$, is furthest from $(0,t)$, then the entire ellipse must be at least as close as this point. In other words, the ellipse will lie entirely inside the circle centered at $(0,t)$ and passing through $(0,-b)$ (radius $t+b$). This will be the case if the radius of curvature of the ellipse at $(0,-b)$ (which is $\dfrac{a^2}{b}$) is less than the radius of curvature of the circle (which is $t+b$). The picture below shows ellipses (possibly ellipse-like ovals) of each case. (Note that for the larger ellipse, the furthest points are not where they intersect the circle - they would be the points where a larger circle is tangent to the ellipse, and it's clear those would not be at the left and right endpoints of the ellipse.

A: Draw pictures, one for $a$ nearly of the size of $b$, one with large $a$ and small $b$ and one with small $a$ and large $b$. Depending on how the ellipse bulges the minimum and maximum will be either on the $y$ axis or the maximum will be close to (but not on) the $x$-axis (the minimum will always be on the $y$-axis). This remark is referring to the absolute maximum. The two 'other solutions' you get may coincide with this absolute maximum or be local maxima, depending on which of the above situations occur.
(Just think of a cigar. If it s 'parallel' to the y-axis, then one end of the cigar is far away from $(0,t)$. If it is parallel to the x-axis, both ends will be far away from $(0,t)$).
A: Yes, there is a geometric way to decide whether the point $(0,-b)$ is at maximum distance from $(0,t)$. Note that this is not always true but that it depends on the shape of the ellipse and the value of $t$.
Just draw a circle with centre $(0,t)$ of radius $r=t+b$. 
Since the circle is the locus of points of equal distance $r$ to the center,  it intersects the ellipse only at $(0,b)$ if and only if the point $(0,-b)$ is at maximal distance to $(0,t)$ Otherwise the circle will have two additional points of intersection with the ellipse. This reasoning assumes and is correct onyl if $(0,t)$ is outside the ellipse.
