Finding the minimum value of a function in an ellipse So I have this problem for my homework:

Consider the ellipse: $\dfrac {x^2}{a^2} + \dfrac{y^2}{b^2}=1$ where $0<b<a$. For every point $(x, y)$ on the ellipse find the the perpendicular line to the ellipse so that the point $(x, y)$ is on that line. This line cuts the ellipse in another point $(x', y')$. Prove that the distance between these two points is $$D(x, y)=\dfrac {2\left(\dfrac {x^2}{a^4} + \dfrac{y^2}{b^4}\right)^{3/2}}{\dfrac {x^2}{a^6} + \dfrac{y^2}{b^6}}\ .$$ 

So I've already done that; but after that it says:

Use Lagrange Multipliers to minimize the function $D(x, y)$.

But the equations get complicated and messy and I don't know if I have to consider the line or just the ellipse. Logically the minimum is at $(0, b)$. Can somebody help me?

I think it is not necessary to use Lagrange Multipliers. we can replace $\dfrac{x^2}{a^2}$ with $1- \dfrac{y^2}{b^2}$, then we have:
$D(y)=2\dfrac{(\dfrac{1}{a^2}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4})^\frac{3}{2}}{\dfrac{1}{a^4}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4}}=2\dfrac{(C_1+C_2y^2)^\frac{3}{2}}{C_3+C_4y^2}$, and $C_1=\dfrac{1}{a^2},C_2=\dfrac{1}{b^2}(\dfrac{1}{b^2}-\dfrac{1}{a^2})=\dfrac{a^2-b^2}{a^2b^4}$,$C_3=\dfrac{1}{a^4},C_4=\dfrac{1}{b^2}(\dfrac{1}{b^4}-\dfrac{1}{a^4})=\dfrac{a^4-b^4}{a^4b^6}$, let $z=y^2$,we have a very simple fomula:
$D(z)=2\dfrac{(C_1+C_2z)^\frac{3}{2}}{C_3+C_4z}$
$D'(z)=2*(\dfrac{3}{2}C_2\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}-C_4\dfrac{(C_1+C_2z)^\frac{3}{2}}{(C_3+C_4z)^2})$
$=2*\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}*(\dfrac{3C_2}{2}-C_4*\dfrac{C_1+C_2z}{C_3+C_4z})$,
since $\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}>0$, when $D'(z)=0$, we have:
$\dfrac{3C_2}{2}-C_4*\dfrac{C_1+C_2z}{C_3+C_4z}=0$, then we put all staff in, we get:
$D'(z)=F(z)*Q(z), F(z)>0$,$Q(z)=b^6-2a^2b^4+(a^4-b^4)z$ , if $D'(z)=0$,then $Q(z)=0$, 
$z=\dfrac{b^4(2a^2-b^2)}{a^4-b^4}$, now here is a trick:
$z=y^2 \leq b^2$ then $\dfrac{b^4(2a^2-b^2)}{a^4-b^4} \leq b^2$,ie.$a^2 \geq 2b^2$ it means only under this condition, $D'(z)=0$ can be satisfied.
when $a^2 \geq 2b^2$, we have:
$Q(z)=b^4(b^2-z)+a^2(a^2z-2b^4), $when $z=b^2, Q(z) \geq 0$ ie $D'(z)\geq 0$;,when $ z=0, Q(z)=b^4(b^2-2a^2) < 0$,ie $D'(z)<0$ , which means D(z) has min. put $z=\dfrac{b^4(2a^2-b^2)}{a^4-b^4}$,we get:
$ D_{min}(z)=\dfrac{\sqrt{27}a^2b^2}{(a^2+b^2)^\frac{3}{2}} \leq 2b $
when $a^2 < 2b^2$, we have:
$Q(z)=a^2z(a^2-2b^2)+b^2(2a^2-b^2)(z-b^2)<0$, which mean when $z=b^2$, $D(z)$ has its min which is $2b$.
BTW, this  is a Japanese Temple Geometry problem which discussed about 300 years ago.
