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?

$\frac{x^2}{4} + \frac{y^2}{9}$ 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. 
