<s>[INST] Distance between ellipse and line What is the distance between the ellipse $$\frac{x^2}{9}+\frac{y^2}{16}=1$$ and the line $y=6-x$.
I think I need to be using Lagrange Multipliers but don't know how.
 [/INST] The minimum value it's the minimal distance between $x+y=6$ and $x+y=c$, 
where $x+y=c$ is a tangent line to our ellipse 
and since $\frac{xx_1}{9}+\frac{yy_1}{16}=1$ is a tangent line to the ellipse in $(x_1,y_1)$,
we obtain $\frac{x_1}{9}=\frac{1}{c}$ and $\frac{y_1}{16}=\frac{1}{c}$. 
We can assume that $c\neq0$ because it's obvious that for $c=0$ we can non get a minimal value.
After using these substitutions to the equation of the ellipse we get for $(x_1,y_1)$:
$\left(\frac{9}{5},\frac{16}{5}\right)$ or $\left(-\frac{9}{5},-\frac{16}{5}\right)$.
The distance between $(x_1,y_1)$ and $x+y=6$ it's $\frac{|x_1+y_1-6|}{\sqrt2}$ and
we see that $\left(\frac{9}{5},\frac{16}{5}\right)$ gets a minimal distance and $\left(-\frac{9}{5},-\frac{16}{5}\right)$ gets a maximal distance. 
Thus, the answer is $\frac{|\frac{9}{5}+\frac{16}{5}-6|}{\sqrt2}$ or $\frac{1}{\sqrt2}$
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