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).
 A: Since focal distance $ = e \times$distance from directrix ($e=$eccentricity = $\frac{\sqrt{a^2-b^2}}{a}$), we only need to locate points on the ellipse that are closest to a directrix and those would be the corresponding end point of the major axis in this case $(a,0)$
A: 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,
$$d=a-\sqrt{a^2-b^2}.$$
A: The hint.
We need to find a a values of $d>0$, for which the system 
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and
$$(x-c)^2+y^2=d^2$$ has solutions.
We obtain:
$$\frac{x^2}{a^2}+\frac{d^2-(x-c)^2}{b^2}=1$$ or
$$\left(\frac{1}{a^2}-\frac{1}{b^2}\right)x^2+\frac{2cx}{b^2}+\frac{d^2-c^2}{b^2}-1=0$$ or
$$-\frac{c^2x^2}{a^2b^2}+\frac{2cx}{b^2}+\frac{d^2-a^2}{b^2}=0$$ or
$$c^2x^2-2ca^2x+a^2(a^2-d^2)=0$$ or
$$(cx-a^2)^2=a^2d^2,$$ which gives
$$x=\frac{a^2+ad}{c},$$ which is impossible or
$$x=\frac{a^2-ad}{c}.$$
Now, we need that the equation $$\frac{\left(\frac{a^2-ad}{c}\right)^2}{a^2}+\frac{y^2}{b^2}=1$$ has real roots for which we need
$$1-\frac{\left(\frac{a^2-ad}{c}\right)^2}{a^2}\geq0$$ or
$$c^2-(a-d)^2\geq0,$$  which gives
$$a-c\leq d\leq a+c.$$
Id est, $d\geq a-c$.
The equality occurs for the point $(a,0)$ on the ellipse, which says that $a-c$ it's the answer. 
Another way.
Let $F_1(c,0)$, $F_2(-c,0)$, $A(a,0)$ and let $M(x,y)$ be a point on the ellipse. 
Thus, $$MF_1+MF_2=2a$$
and by the triangle inequality $$MF_1+F_1F_2\geq MF_2$$ or
$$2MF_1+F_1F_2\geq MF_1+MF_2$$ or
$$2MF_1+2c\geq2a,$$ which gives
$$MF_1\geq a-c.$$
The equality occurs, when $M\equiv A$, which says that $a-c$ is a minimal value.
