So I have this problem for my homework:
Consider the ellipse: $\dfrac {x^2}{a^2} + \dfrac{y^2}{b^2}=1$ where $0. 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?