Find shortest distance from point on ellipse to focus of ellipse. 
Find the point $(x,y)$ on the ellipse $b^2x^2 + a^2y^2 = a^2b^2$ such that the distance to the focus $(c,0)$ is a minimum.

My book I got this problem out of gave a suggestion saying to express the distance as a function of $x$ and work the problem and then express the distance as a function of $y$ and work the problem.
So, I tried to work it out:
$$x^2 = \frac{a^2b^2 - a^2y^2}{b^2}$$
$$y^2 = \frac{a^2b^2 - b^2x^2}{a^2}$$
Now, I write the distance function:
$$d = \sqrt{(x-c)^2 + y^2}$$
Then, I express the distance function as a function of x:
$$d = \sqrt{(x-c)^2 + \frac{a^2b^2 - b^2x^2}{a^2}}$$
Then, I took the derivative w.r.t $x$:
$$d' = 2x - 2c - \frac{2b^2}{a^2}$$
Now, I did as the book told me, and expressed the distance function as a function of y:
$$d' = -\frac{2a^2}{b^2} + \frac{4a^2c}{bx} + 2y$$
I tried to set both derivatives to zero and attempted to solve, but I couldn't really figure out what to do (my algebra is poor).
Any help would be appreciated.
Thanks.
Edit: I need a calculus solution (I'm working out of a calculus book).

Assuming $a>b$, we have $c=\sqrt{a^2-b^2}$.
Using the parametric equations
$$\begin{cases}x=a\cos\theta,\\y=b\sin\theta\end{cases}$$
we minimize
$$(a\cos\theta-c)^2+(b\sin\theta)^2.$$
The derivative of this expression is
$$-a\sin\theta(a\cos\theta-c)+b\cos\theta(b\sin\theta).$$
There is a root for $$\sin\theta=0$$ (on the major axis) and for
$$(a^2-b^2)\cos\theta=ac,$$ or 
$$\cos\theta=\frac a{\sqrt{a^2-b^2}}>1 (!).$$
Hence,
$$d=a-\sqrt{a^2-b^2}.$$
