I need to find the distance between the ellipse $\epsilon = \{(x,y):\tfrac{x^2}{4}+\tfrac{y^2}{9}=1\}$ and the line $D = \left\{(x,y):5x-3y=17\right\}$ .
I know that I have to find the nearest tangent line which is parallel to $D$, and like that, I can find the distance, which is $\frac{\sqrt{181}-17}{\sqrt{34}}$.
The problem is that I can't prove why the tangent line must be used. Why the closest point must be on it?
On
The underlying reason you need the tangent that is parallel to the line of concern is because you really want the line that is perpendicular to that line, and goes through a point on that ellipse.
So, this is the setup of your curves:
In the end, you want to measure distance, so you need two things, which are very interrelated:
If you think about playing with various different perpendicular lines, you'll see one gives the other. Note that your line is $$ 5x - 3y = 17 \implies y = \frac 5 3 x - \frac{17}{3} $$ so the perpendicular line $y_\perp$ takes the form $$ y_\perp = - \frac 3 5 x + b $$ for some undetermined $b$. This $b$ can be found in various ways.
Here, I've generated $y_\perp$ for each $b \in \{-5,-4,\cdots,4,5\}$:
Notice how that the point on the ellipse closest to the line $y$ is on one of the more-middle lines, where the ellipse bulges towards the line.
Well, for this closest point, the length of the green line segment from that point to the black line would be the distance. And all of the points that are that distance will lie on a line $y_\parallel$ parallel to the original as a result (because it is perpendicular to $y_\perp$).
(That is to say, if $\ell \parallel m$, then each point on $\ell$ is a fixed distance away from the line $m$, and each point on $m$ is a fixed distance from the line $\ell$, and you can find that distance with the perpendicular line through both.)
Of course, $y_\parallel$ takes the form $$ y_\parallel = \frac 5 3 x + c $$ for some undetermined constant $c$. Graphing some for $c \in \{-6,-4,\cdots,2,4\}$ in purple, we get the (cluttered) diagram,
But in particular, if one envisions twiddling $b$ and $c$ a bit, we get something like this:
The relevance of the tangency to the entire process becomes immediately clear.
Of course, one should be wary that multiple parallel lines to $y$ may arise, as you can see here:
I've made a Desmos demo here you can play with, to convince yourself of the relevant details by playing with some free parameters.
In particular, consider trying to move $y_{\text{parallel}}$ in the demo until it touches one and only one point of the ellipse: this is the point of tangency. Then you just need the perpendicular line, $y_{\text{perpendicular}}$, to go through that point. The distance between that tangency point, and where the perpendicular line meets the original line, is the distance desired.
(Note that the answers given in the demo will not necessarily be exact, just approximations based on how a graph looks!)
Two lines in the plane cross if they are not parallel. Given two points on those lines, we can reduce the distance between them by moving closer to the point where the lines cross.
In your situation, if the tangent line to a point on the ellipse is not parallel, then moving the point along the ellipse changes the distance to the line. So, at the minimum distance, the tangent line must be parallel to the given line.