I want to find out ${(u,v)}$ on the ellipse $$\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$$ for a point ${(x,y)}$ inside ellipse, which will denote shortest distance from $(x,y)$ to the ellipse boundary. I expressed this problem as a constrained optimization problem by this Lagrangian:
\begin{align} L(u,v,\lambda) &= (u-x)^2+(v-y)^2-\lambda\left(\frac{u^2}{a^2} + \frac{v^2}{b^2}-1\right)\\[1ex] \frac{\partial L(u,v,\lambda)}{\partial u}&=2u-2x-\frac{2\lambda u}{a^2}\\[1ex] \frac{\partial L(u,v,\lambda)}{\partial v}&=2v-2y-\frac{2\lambda v}{b^2}\\[1ex] \frac{\partial L(u,v,\lambda)}{\partial \lambda}&=\frac{u^2}{a^2}+\frac{v^2}{b^2}-1\\[1em] u&=\frac{a^2x}{a^2-\lambda}\tag1\\[1ex] v&=\frac{b^2y}{b^2-\lambda}\tag2\\[1ex] \frac{u^2}{a^2}&+\frac{v^2}{b^2}-1=0\tag3\\ \end{align}
After substituting $u$ and $v$ in equation $(3)$, I get \begin{aligned} \frac{a^2x^2}{(a^2-\lambda)^2}+\frac{b^2y^2}{(b^2-\lambda)^2}-1=0\\ \end{aligned} I am getting a 4th degree equation of $\lambda$. I am facing difficulties in finding algebraic solution of the equation.
