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.
 A: 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$=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=\dfrac{12\sqrt{7}}{7}$
A: The tangent line to an ellipse parallel to a given line can be found without using calculus.
First, notice that finding the distance between a circle $C$ and a line $L$ would be a simpler problem. One would find line $L^\prime\perp L$ that passes through the center of the circle and then find the intersection point of $C$ and $L^\prime$ on the side of $C$ nearest $L$. Then one could find the distance from this point to the line.
Even though we have a circle rather than an ellipse, an ellipse can be transformed into a circle by either stretching it along its minor axis or shrinking it along its major axis. And the nice thing about stretching and shrinking transformations is that, not only do such transformations transform lines onto lines, they, in fact, transform parallel lines onto parallel lines!
The ellipse $\dfrac{x^2}{4}+y^2=1$ can be transformed onto the circle $x^2+y^2=1$ by the shrinking transformation
$$T\,:\,x\mapsto 2x$$
This transformation maps the given line $L: y=-\dfrac{\sqrt{3}}{2}x+8$ onto the line $T(L):y=-\sqrt{3}x+8$. Any line perpendicular to $T(L)$ has slope $\dfrac{1}{\sqrt{3}}$. Since we want the line perpendicular to $T(L)$ passing through the center, which in this case is the origin, we want the line $y=\dfrac{1}{\sqrt{3}}x$. Substituting this value of $y$ into the circle equation gives their point of intersection lying nearest the line $T(L)$ as $\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)$.
Next we are going to find the equation of the line tangent to the circle at the point $\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)$. We know it has slope $-\sqrt{3}$ because it is parallel to $T(L)$. So we easily find that its equation is $y=-\sqrt{3}x+2$.
Now we apply the inverse transformation
$$T^{-1}\,:\,x\mapsto\dfrac{1}{2}x$$
to this line to obtain $y=-\dfrac{\sqrt{3}}{2}x+2$. This line is the $L^\prime$ which is tangent to the ellipse and parallel to $L$. Furthermore, it contains the point of tangency $T^{-1}\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)=\left(\sqrt{3},\dfrac{1}{2}\right)$ which is the nearest point of the ellipse to $L$.
Then the distance is found using the distance formula to be exactly what you found, $\dfrac{12\sqrt{7}}{7}$.
No calculus required.
A: 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 $-\sqrt{a^2+b^2}\le a\cos t+b\sin t\le\sqrt{a^2+b^2}$
