Suppose you have the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Suppose somewhere on the interior you have point $P(m,n)$. What is the closest point on the ellipse to the $P$? Can this be found without having to solve a 4th degree polynomial?
Because of symmetry, we can assume that $P$ is in the First Quadrant or on its boundary without loss of generality.
Lagrange multipliers come to mind. Perpendicularity. Figure most efficient approach is parameterization.
Let $x=a\cos t$, $y=b\sin t$. Let $d^2=(x-m)^2+(y-n)^2$.
Take the derivative of $d^2$ and set it to zero.
$$0=2(a\cos t-m)(-a\sin t)+2(b\sin t-n)(b\cos t)$$ $$0=(x-m)(\frac{-a}{b}y)+(y-n)(\frac{b}{a}x)$$
This represents a hyperbola passing through the origin and containing $P(m,n)$.
Expanding, then rearranging:
$$0=-2a^2\sin t\cos t+2ma\sin t+2b^2\sin t\cos t-2nb\cos t$$ $$nb\cos t=(b^2-a^2)\sin t\cos t+ma\sin t$$
Squaring both sides: $$n^2b^2\cos^2t=(b^2-a^2)^2\sin^2 t\cos^2 t+m^2a^2\sin^2 t+2ma(b^2-a^2)\sin^2 t\cos t$$
That $\cos t$ appears to the first power and $\sin t$ only to the second, this suggests swapping out $\sin^2 t$ with $(1-\cos^2 t):
$$n^2b^2\cos^2t$$ $$=(b^2-a^2)^2\cos^2t(1-\cos^2 t)+m^2a^2(1-\cos^2t)-2ma(b^2-a^2)(1-\cos^2t)\cos t$$ or,
$$n^2b^2\cos^2t=$$ $$(b^2-a^2)^2\cos^2t -(b^2-a^2)^2\cos^4t+m^2a^2-m^2a^2\cos^2t +2ma(b^2-a^2)\cos t -2ma(b^2-a^2)\cos^3t$$
Isolating $\cos t$ to one side: $$(b^2-a^2)^2\cos^4t+2ma(b^2-a^2)\cos^3t+[n^2b^2-(b^2-a^2)^2+m^2a^2]\cos^2 t-2ma(b^2-a^2)\cos t-m^2a^2=0$$
We now have a 4th order polynomial in $\cos t$. Numerical methods are available, but can it be reduced further?
I think I've made some progress. The x-y form of the equation to be solved represents a hyperbola that passes through the origin as P(m,n). So there must be at least 2 points of intersection of the hyperbola with the ellipse.
Swapping in $(z+1/z)/2=\cos t$ and $(z-1/z)/(2i)=\sin t$ and introducing $z_0$ to simply the coefficients a complex equation comes up: $$z^4+2z_0z^3-2\bar{z_0}z-1=0$$