The Path of the center of a cutting tool that will cut out an ellipse (x/4)2+(y/2)2=1 I am currently stuck on this question and it's implementation on Mathematica.
You are the chief engineer at the Badger Steel Plate Company in Madison, Wisconsin. In comes an order for 750 square steel plates, each measuring 12 inches wide and 12 inches long.
Go to the drawing board
plate = Graphics[{Blue, Thickness[0.01], Line[{{-6, -6}, {-6, 6}, {6, 6}, {6, -6}, {-6, -6}}]}]; 
plateplot = Show[plate, Axes -> True, AxesOrigin -> {0, 0}, AspectRatio -> Automatic, AxesLabel -> {"x", "y"}]

But here's the kicker:
The plates are to have everything inside the ellipse $(x/4)^2+(y/2)^2=1$ cut out.
Take a look:
Clear[x, y, t]; 
x[t_] = 4 Cos[t]; 
y[t_] = 2 Sin[t]; 
hole = ParametricPlot[{x[t], y[t]}, {t, 0, 2 Pi}, PlotStyle -> {{Blue, 
Thickness[0.01]}}]; 
Show[plateplot, hole]


You have a new robotic router that takes instructions from Mathematica and whose cutting center can be programmed to follow any curve you tell it to follow. If you are going to use a bit 1 inch in diameter, then what curve should you program in as the path of the center of the router to cut out the ellipse?
After you have found the correct curve, add its plot to the plot above.
Big Tip:
If your curve is an ellipse, then you screwed up.
The normal vector D[unittan[t], t] could be very useful.
I was able to obtain the plot using normal vector D[unittan[t], t] for the plot which produced the Following plot
Now I have the following question.
Actually, the bit size of 1 inch in diameter used above was arbitrarily chosen by reaching into the 
drawer and pulling out a bit. You could always get by with a smaller bit. Why?
But you cannot use bits that are too large. Why?
Try to estimate the diameter of the largest bit that you could use to do the job.

Any help on demystifying this question will be greatly appreciated
 A: The largest bit you can use has the size of the osculating circle at the pointier vertices of the ellipse. Any larger bit will not be able to cut the region around those vertices; smaller bits always work because the parallel curves that those bits travel on to cut out the ellipses are smooth.
The exact osculating radius at the points $(\pm4,0)$ is, according to a formula conveniently placed on Wikipedia, $\frac{2^2}4=1$. Thus the largest bit that can drill the given ellipse is $2$ inches wide.
A: Write the equation of the ellipse in polar coordinates -
$r(\theta) = \displaystyle \frac{ab}{\sqrt{b^2 \cos^2\theta + a^2 \sin^2 \theta}}$, Center at $(0,0)$ and $(0 \le \theta \le {2\pi})$
Here, major axis $a = 4$, minor axis $b = 2$.
So, $r(\theta) = \displaystyle \frac{4}{\sqrt{1 + 3 \sin^2 \theta}}$
Now if the radius of the bit is $r_b$, what is the path of the point on the bit which is closer to the origin? The curve that it will follow from the origin $(0,0)$ over $0 \le \theta \le {2\pi}$ will be the radius of the ellipse - the diameter.
$\displaystyle r_c(\theta) = \frac{4}{\sqrt{1 + 3 \sin^2 \theta}} - 2r_b = \frac{4 - 2r_b{\sqrt{1 + 3 \sin^2 \theta}}}{\sqrt{1 + 3 \sin^2 \theta}}$ ...(i)
So for $r_b = 0.5$ unit (dia = 1 unit), please see below. I have shown $3$ curves - one is the ellipse, the middle one is the one that center of the bit follows (instead of $2r_b$, use $r_b$ in equation (i)) and the inside one is the one that the inner side of the bit follows.

As we increase $r_b$ and it reaches $1$ (dia reaches 2 which is the minor axis of ellipse), it will touch the center (origin) at $\theta = \pi/2$ and $3\pi/2$. You can see $r_c(\pi/2) = 0$ from equation (i). You cannot have $r_b \gt 1$ (a bit of dia bigger than 2 unit) as $r_c$ (radius for the path followed by inner side of the bit) will have a negative value at certain angles.
See below for $r_b=1$. Again $3$ curves as explained earlier.

A: I use differential calculus here.
$$ (x/a)^2+(y/b)^2=1$$
Differentiate once wrt $x$
$$ \tan \phi= dy/dx ,\; \frac{x}{a^2}+ \frac{y \tan \phi}{b^2}=0$$
Again differentiate wrt arc, with which it is
primed. Differentials of arc along the axes are
$$ (dx,dy)=ds (\cos \phi, \sin \phi)$$
ODE of elliptic arc is
$$ \phi'(s) y(s)\cdot \sec\phi + (\cos(\phi(s) \;b/a)^2 =0 $$

Cutter center location on red "parallel ellipse" produces axes $(a,b)$ $= (48\times 24) $ inches in blue. Dimension normal to elliptic arc is always constant.
A milling cutter bit radius $rmc= 3 $ inches is used but can be changed. Table of values on red curve can be generated from Mathematica listing given. Outer milled profile of cut channel is also sketched.
Clear["Global`*"]
phi=0.+10^-7;xi=0.;smax=24;a=4;b=2;rmc=-3/12;yi=b;
eq={PH'[s]Sec[PH[s]]Y[s]+Sin[PH[s]]^2+(Cos[PH[s]] b/a)^2==0,PH[0]==phi,X'[s]==Cos[PH[s]],Y'[s]==Sin[PH[s]],Y[0]==yi,X[0]==xi};
NDSolve[eq,{PH,X,Y},{s,0,smax}];
{ph[t_],x[t_],y[t_]}={PH[t],X[t],Y[t]}/.First[%];
m1=ParametricPlot[{x[s],y[s]},{s,.0,smax},GridLines->Automatic,AspectRatio->Automatic,PlotStyle->{Blue,Thick}];
m2=ParametricPlot[{x[s]-rmc Sin[ph[s]],y[s]+rmc Cos[ph[s]]},{s,.0,smax},PlotLabel->EllipseBlank,GridLines->Automatic,AspectRatio->Automatic,PlotStyle->{Red,Thick}];
Show[{m1,m2},PlotRange->All]

