TITLE: Shortest distance from ellipse to line
QUESTION [5 upvotes]: What is the shortest distance between the ellipse 
$$\frac{x^2}{4}+y^2=1$$
and the line $y=\frac{-\sqrt{3}}{2}x+8?$
I tried solving it by using a line that is tangent to the ellipse and parallel to the line given: $y=\frac{-\sqrt{3}}{2}x+b$
Then plugging the y back into the ellipse equation: 
$x^2-b\sqrt{3}x+(b^2-1)=0$ 
Which means the discriminant would have to equal 0 for the line to be tangent, right? 
$3b^2-4b^2+4=0$
$b=2$
But when I plug the y-intercept, (0,2), into a point to line distance formula I get 
$$\frac{|0*\sqrt{3}+2*2-16|}{\sqrt{3+4}}$$
Which is $\frac{12 \sqrt{7}}{7}$, but the answer key says it's $\frac{3\sqrt{13}}{2}$
What am I doing wrong? I don't mind learning any other methods, but I haven't learned any calculus yet, so I would like to do the problem without it. Thank you.

REPLY [2 votes]: WLOG any point (P) on the ellipse can be chosen to be $(2\cos t,\sin t)$
What's is the perpendicular distance of $P$ from the given line?
Finally use $-\sqrt{a^2+b^2}\le a\cos t+b\sin t\le\sqrt{a^2+b^2}$