Shortest distance between ellipse and a line I was trying to find the shortest distance between the ellipse
$$\frac{x^2}{4} + y^2 = 1$$
and the line $x+y=4$. We have to find the point on the ellipse where 
its tangent line is parallel to $x+y=4$ and find the distance between those two points. 
However, when I used the implicit differentiation, I get 
$$\frac{x}{2} + 2y\frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = \frac{-x}{4y}$$
If it's parallel to $x+y=4$, then we need $x=4y$. Do I just plug it into ellipse equation and solve for it and calculate the distance between the point and a line or am I doing it wrong? I just wanted to clarify. Any help would be appreciated. Thanks!

Simpler solution:
Scale the horizontal axis such that $u=\frac{x}{2}$. Then, the ellipse equation becomes:
$$u^2+y^2=1$$
Likewise, the equation of the line becomes:
$$2u+y=4$$
Now we simply must find the closest distance between that line and the unit circle. This can be done fairly easily using analytic geometry. First find the equation of a line that goes through the origin and is perpendicular to the other line:
$$y=\frac{1}{2}u$$
Now find the intersection point of the two lines. I.e., solve the system:
$$2u+y=4$$
$$y=\frac{1}{2}u$$
The solutions are $u=\frac{8}{5}$ and $y=\frac{4}{5}$. The second line intercepts the unit circle at $u=\frac{2}{\sqrt{5}}$ and $y=\frac{1}{\sqrt{5}}$ (trigonometry not even required – use simple substitution).
So the shortest distance between the circle and the line is the distance between $(\frac{8}{5},\frac{4}{5})$ and $(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})$. If we rescale our coordinate plane by substituting back in: $u=\frac{x}{2}$, then we will find the shortest distance between our ellipse and line as being simply the distance between $(\frac{16}{5},\frac{4}{5})$ and $(\frac{4}{\sqrt{5}},\frac{1}{\sqrt{5}})$. This is simply given by:
$$\sqrt{\Big(\frac{16}{5}-\frac{4}{\sqrt{5}}\Big)^2+\Big(\frac{4}{5}-\frac{1}{\sqrt{5}}\Big)^2}=\boxed{\frac{\sqrt{17(21-8\sqrt{5})}}{5}}$$
No calculus required!
