How to find the minimum/maximum distance of a point from elipse
I have the point $(1,-1)$
and the ellipse $$x^2/9 + y^2/5 = 1 $$
How to find the minimum and maximum distance of the point from the ellipse ?
from exploring the ellipse I know that $$a = 3$$ , $$b =\sqrt{5}$$ $$ c = \sqrt{a^2-b^2} =\sqrt{9-5} = \sqrt{4}=2$$
the eccentricity of the ellipse is $$e=c/a = 2/3 $$ the center is $(0,0)$ and the guides are $$x=3,~~x=-3,~~y=\sqrt{5},~~y=-\sqrt{5}$$
the focus points are : $(2,0)$ and $(-2,0)$
how from all of that do I find the requested in the question?
Using $Mathematica$, I have plotted the solution, which corresponds to roots of a quartic equation, which is why I am only going to show you a picture and a numerical approximation of the coordinates, which are $(1.38065, -1.98519)$, and $(-2.84987, 0.698515)$.
The ellipse can be parametrized as follows: $\alpha(t) = \langle 3\cos(t), \sqrt{5}\sin(t)\rangle$ such that $0 \leq t \leq 2\pi$.
From here, note that finding the points that minimize and maximize the distance will be the same points that minimize/maximize the square of the distance. With this trick, we can eliminate some yucky square roots. Applying the Pythagorean theorem, we can define a function $f$ that represents the square of the distance from $(1, -1)$ to an arbitrary point on the ellipse:
$$f(t) = \Big(1 - 3\cos(t)\Big)^2 + \Big(-1 - \sqrt{5}\sin(t)\Big)^2$$
Computing the derivative of this function, we get:
$$f'(t) = 2\cos(t)\Big(\sqrt{5} - 4\sin(t)\Big) + 6\sin(t)$$
The derivative has $2$ zeros on the interval $[0, 2\pi]$. Those should be the $t$-values that minimize and maximize the distance from your point.