<s>[INST] Shortest distance from ellipse to line 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.
 [/INST] I think there is a typo in the answer.
Note that the shortest distance is always perpendicular distance.
Let $P$ be a point parallel to the given line $y=\frac{-\sqrt{3}}{2}x+8$.
Then the slope at $P$ is 
$$\frac{dy}{dx}=-\frac{x}{4y}\mbox{------(1)}$$
Since the given line and tangent line are parallel, the slopes are equal.
$$\mbox{Slope of the given line is } \sqrt{3}x+2y=16 \mbox{ is }\frac{-\sqrt{3}}{2}$$
$$-\frac{x}{4y}=\frac{-\sqrt{3}}{2}$$
$$x=\sqrt{3}y$$
Now substitute the $x=\sqrt{3}y$ in the ellipse $x^2+4y^2=4$ and we get
$$12y^2+4y^2=4$$
$$y=\pm\frac12$$
Clearly $\left(\sqrt{3},-\dfrac12\right)$  is closer to the line.
Distance$=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=\dfrac{12\sqrt{7}}{7}$
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