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.
Just another approach, for the sake of variety.

You can also observe that at the point of closest approach, the ellipse must be parallel to the line $y = 6-x$, which has slope $-1$.  Implicit differentiation of the ellipse gives us

$$
\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx} = 0
$$

$$
16x+9y\frac{dy}{dx} = 0
$$

$$
\frac{dy}{dx} = -\frac{16x}{9y}
$$

which must equal $-1$ at the point of closest approach (with $x, y &gt; 0$—a quick sketch will show why), so

$$
y = \frac{16x}{9}
$$

Plug this back into the equation for the ellipse, solve for the point, and obtain the distance.
Yes, you can use Lagrange multipliers minimize $x^2+ y^2$ (which is equivalent to minimizing $\sqrt{x^2+ y^2}$) subject to the constraints $\frac{x^2}{9}+ \frac{y^2}{16}= 1$ and $x+ y= 6$. $\nabla x^2+ y^2= &lt;2x, 2y&gt;$, $\nabla \frac{x^2}{9}+ \frac{y^2}{16}= &lt;2x/9, y/8&gt;$, and $\nabla x+ y= &lt;1, 1&gt;$.

So you have $&lt;2x, 2y&gt;= \lambda&lt;2x/9, y/8&gt;+ \mu&lt;1, 1&gt;$ together with the constraints to solve for x and y, $\mu$, and $\lambda$.