<s>[INST] 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?
 [/INST] The lack of symmetry in the geometric arrangement doesn't work in our favor, but fortunately the coefficients in the equations are only unpleasant to work with near the very end.  As indicated by Padmanabha P Simha, we can use the Lagrange-multiplier method to aid in starting our work.  We can extremize the "distance-squared" function, as Kaj Hansen observes, since the distance from $ \ (1, \ 1) \ $ to a point $ \ (X, \ Y) \ $ on the ellipse can only be non-negative.  We will derive the constraint function from the equation for the ellipse as
$$ \frac{x^2}{9} \ + \ \frac{y^2}{5} \ = \ 1 \ \ \rightarrow \ \ g(x,y) \ = \ 5x^2 \ + \ 9y^2 \ - \ 45 \ \ . $$
For the distance-squared function $ \ f(x,y) \ = \ (x - 1)^2 \ + \ (y + 1)^2 \ $ , the Lagrange equations are
$$ \frac{\partial f}{\partial x} \ = \ \lambda \ \cdot \ \frac{\partial g}{\partial x} \ \ \Rightarrow \ \ 2 \ (x - 1) \ = \ \lambda \ \cdot \ 10x \ \ , $$
$$ \frac{\partial f}{\partial y} \ = \ \lambda \ \cdot \ \frac{\partial g}{\partial y} \ \ \Rightarrow \ \ 2 \ (y + 1) \ = \ \lambda \ \cdot \ 18y \ \ . $$
The result for the multiplier is $ \ \lambda \ = \ \frac{1}{5} \ \frac{x - 1}{x} \ = \ \frac{1}{9} \ \frac{y + 1}{y} \ $ .  (We will make use of this relation shortly.)
Alternatively, we can show (for instance, by applying Lagrange multipliers) that the segments from a point external to a curve to an extremal point of the curve are perpendicular (normal) to the curve.  If we differentiate the equation for the ellipse implicitly with respect to $ \ x \ $ , we obtain
$$ \frac{d}{dx} \ [ \ 5x^2 \ + \ 9y^2 \ ] \ = \ \frac{d}{dx} \ [45] \ \ \Rightarrow \ \ \frac{dy}{dx} \ = \ - \frac{5}{9} \ \frac{x}{y} \ \ , $$
giving us the slope of a tangent line to a point on the ellipse.  The slope of a normal line at that point is then  $ \  \frac{9}{5} \ \frac{y}{x} \ $ .  The line segment from $ \ ( 1, \ 1) \ $ to that point has a slope of $ \ \frac{y - (-1)}{x - 1} \ $ , so we find  $ \ \frac{y + 1}{x - 1} \ = \ \frac{9}{5} \ \frac{y}{x} \ $ , which is equivalent to our Lagrange result (since this is what the method locates for us, as nicely illustrated by heropup's tangent circles).
We can re-arrange this equation as  $ \ y \ = \ \frac{5x}{4x - 9} \ $ , which is a hyperbola.  Upon graphing this along with our ellipse, we see that the two curves intersect where the local normals correspond to the lines also passing through $ \ ( 1, \ 1) \ $; the added rays indicate those normal lines.

We need to locate those intersections, so we may insert our expression for $ \ y \ $ into the ellipse equation to produce
$$ 5x^2 \ + \ 9 \ \left( \ \frac{5x}{4x - 9} \ \right)^2 \ - \ 45 \ = \ 0 \ \ \Rightarrow \ \ x^2 \ - \ 9 \ +  \ \frac{9 \cdot 5x^2}{(4x - 9)^2}  \ = \ 0 $$
$$ \Rightarrow \   \ \frac{(x^2  -  9)(4x - 9)^2 \ + \ 45x^2}{(4x - 9)^2}  \ = \ 0 \ \ . $$
Since $ \ 4x - 9 \ = \ 0 \ $ is not in the domain of the rational function for $ \ y \ $ , we can simply solve for the values of  $ \ x \ $ at which the numerator of the left-hand-side ratio is zero.  Multiplying out the factors and simplifying leaves us to solve the quartic equation $ \ 16x^4 \ - \ 72x^3 \ - \ 18x^2 \ + \ 648x \ - \ 729 \ = \ 0 \ $ , which I left to WolframAlpha (the exact values for the two real and two complex conjugate solutions are rather horrifying).  The closest and most distant points on the ellipse to $ \ (1, \ -1) \ $  and the minimal and maximal distances are then as given in heropup's answer and comment.
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