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?
Best Answer
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.