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. 

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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 $\theta = \arcsin\left(\frac{bt}{b^2-a^2}\right)$ ?

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Is there an geometry way to see this conclusion : the maximum is obtain at $(0,-b)$? thanks very much