Using Lagrange's multiplier method, find the shortest distance between the line y=10-2x and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ 
Using Lagrange's multiplier method, find the shortest distance between the line $y=10-2x$ and the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.

My work: 
Let the point on ellipse be $(2\cos\theta, 3\sin\theta)$
Let $F=(x-2\cos\theta)^2+(y-3\sin\theta)^2+\alpha(2x+y-10)$.
I partially diffentiated $F$ wrt $x$,$y$ and $\theta$ equated to $0$.
I get $\tan\theta=\frac{3}{4}$ and $\alpha=-2$
The point on ellipse I get is $(\frac{8}{5},\frac{9}{5})$ and the point on line I get is $(\frac {18}{5}, \frac{14}{5})$
Is this method correct? Can I take $\theta$ to be an independent variable, and differentiate wrt to it?
 A: Although your solution method is fine, I suspect that whoever gave you this exercise intended for you to use the equation of the ellipse directly instead of parameterizing the curve. Once you’ve done the latter, you have a single-variable problem for which there’s not much point in using a Lagrange multiplier: you can compute the distance of the parameterized point to the line and minimize that directly.  
The intended solution is likely something along these lines: The square of the distance from a point $(x,y)$ to the line is given by $$f:(x,y)\mapsto{(2x+y-10)^2\over5}.$$ You are to minimize this function subject to the constraint $g(x,y)=0$, where $g:(x,y)\mapsto\frac{x^2}4+\frac{y^2}9-1$, so form the function $F:(x,y,\lambda)\mapsto f(x,y)-\lambda g(x,y)$, set the partial derivatives of $F$ to zero and solve for $x$, $y$, $\lambda$. This will give you critical points of $f$ that you’ll still have to test to see which are minima.
A: I think it should be
$$F=(x-t)^2+(y-s)^2+\alpha(2x+y-10)+\beta\left(\frac{t^2}{4}+\frac{s^2}{9}-1\right),$$
but I think it's better to solve your problem without Lagrange multipliers. 
A: Alternatively: you can find the tangent line to the ellipse that is parallel to the line $y=10-2x$.
The tangent line to the ellipse at the point $(x_0,y_0)$ must have the slope $-2$:
$$\left(\frac{x^2}{4}+\frac{y^2}{9}\right)'_x=(1)'_x \Rightarrow \frac{x}{2}+\frac{2y}{9}y'=0 \Rightarrow y'=-\frac{9x}{4y} \Rightarrow$$
$$y'(x_0)=-\frac{9x_0}{\pm 4\cdot3\sqrt{1-\frac{x_0^2}{4}}}=-2 \Rightarrow x_0=\pm\frac{8}{5} \Rightarrow y_0=\pm\frac95.$$
Note: In general, there can be two tangent lines and $x_0=-\frac85,y_0=-\frac95$ must be ignored as it is the furthest point from the given line (i.e. the max distance).
Then you should find the distance from the point $(\frac85,\frac95)$ of ellipse to the line $y=10-2x$.
A: $\frac{x^2}{4} + \frac{y^2}{9}$ can be made into $ 1 = 9x^2 + 4y^2 = 36 $ which can be made into $9cos^2(\theta) + 4sin^2(\theta) = 36$. This should help simplify your calculations. 
