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!
 A: If $F(x,y) \equiv \frac{1}{4}x^2 + y^2$, then $\nabla F = (\frac{1}{2}x, 2y)$ is orthogonal to curves of constant $F$, hence orthogonal to the ellipse when $(x,y)$ is on the ellipse. Also make $\nabla F$ orthogonal to the given line, so $(\frac{1}{2}x, 2y)\cdot (1, -1) = 0$ gives $y = \frac{1}{4}x$.
A: Another, hopefully correct in the basics, hints:
You want to minimize the distance function
$$\frac{|x+y-2|}{\sqrt2}\;\;<--\;\;\text{distance of a point to line}\;\;x+y-4=0$$
subject to the constraint
$$\frac12x^2+y^2-1=0$$
Putting things this way, we don't really care anymore whether the line intersects or not the ellipse! Now proceed as usual.
A: actually,only rewrite the function of the ellipse as $y=\sqrt{1-\frac{x^2}{4}}$,the upper half part.
$k=\frac{dy}{dx}=\frac{-\frac{x}{2}}{2\sqrt{1-\frac{x^2}{4}}}$,
thus,the point$({x_0},{y_0})$ on the ellipse and $\frac{-\frac{x_0}{2}}{2\sqrt{1-\frac{x_0^2}{4}}}=-1$, solving , we have $x_1=\sqrt{2},x_2=-\sqrt{2}$(not satisfied).
so,$x_0=\sqrt{2}$,and by the ellipse equation,we have $y_0=\frac{1}{\sqrt{2}}$.
THE DISTANCE BETWEEN THE POINT$({x_0},{y_0})$ TO A LINE $x+y=4$ IS TRIVIAL.I HOPE IT IS HELPFUL .
A: 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!
A: Perhaps you can try explicit differentiation. From the ellipse equation:
$$y=\sqrt{1-\frac{x^2}4}$$
so
$$y'=\frac{-x}{4\sqrt{1-\frac{x^2}4}}$$
(If you display the ellipse and the line, it is clear that the point you are looking for is in the upper half of the ellipse).
Then look for the point where this derivative is equal to the slope of the line:
$$\frac{-x}{4\sqrt{1-\frac{x^2}4}}=-1$$
Then solve this equation and it si done.
Arternatively, you can solve the system:
$$\left\{ \begin{array}{rcl}
\frac{x^2}4+y^2&=&1\\
x+y&=&k
\end{array}
\right.$$
When you get a discriminant that depends on $k$, it must be $0$. Find $k$ and then the point where the line touches the ellipse. You will obtain two values for $k$: you should take the greater, since the line is "above" from the ellipse.
A: Any point on the ellipse can be represented as $\displaystyle P(2\cos\phi,\sin\phi)$
So, if $s$ is the distance of $P$ from the given line is $$s=\frac{|2\cos\phi+\sin\phi-4|}{\sqrt{1^2+1^2}}$$
So, we need minimize $\displaystyle|2\cos\phi+\sin\phi-4|$
We can achieve this by Second derivative test.
Otherwise, setting $\displaystyle2=r\cos\psi,1=r\sin\psi\implies \tan\psi=2$ and $r=\sqrt{2^2+1^2}=\sqrt5$
$\displaystyle2\cos\phi+\sin\phi=\sqrt5\sin\left(\phi+\arctan2\right)$
$\displaystyle\implies-\sqrt5\le2\cos\phi+\sin\phi\le\sqrt5$
$\displaystyle\implies-\sqrt5-4\le2\cos\phi+\sin\phi-4\le\sqrt5-4$
$\displaystyle\implies\sqrt5+4\ge4-2\cos\phi-\sin\phi\ge4-\sqrt5$
$\displaystyle\implies\sqrt5+4\ge|4-2\cos\phi-\sin\phi|\ge4-\sqrt5$
$\displaystyle\implies\sqrt5+4\ge|2\cos\phi+\sin\phi-4|\ge4-\sqrt5$
$\displaystyle\implies\frac{\sqrt5+4}{\sqrt2}\ge s\ge\frac{4-\sqrt5}{\sqrt2}$
