The Envelope of Rotating Ellipses Around Their Focus: Part I The  classical ellipse (in polar coordinates)
$$ p/r = 1- \epsilon \, \cos \theta $$ rotates about its focus.
How should $p$ dilate with respect to $\theta$ so as to have one 
cartesian envelope $ x= \dfrac{a}{1-\epsilon}? $
 A: Here's a brute-force approach.
The ellipse rotated by angle $\alpha$ has polar equation
$$p = r - r e \cos(\theta-\alpha) = r - e (r\cos\theta\,\cos\alpha + r\sin\theta\,\sin\alpha) \tag{1}$$
We can transform this into Cartesian coordinates via 
$$r\cos\theta \to x \qquad r\sin\theta \to y \qquad r\to \sqrt{x^2+y^2}$$
to get 
$$x^2 + y^2 = \left(\;p + x e \cos\alpha + y e \sin\alpha\;\right)^2 \tag{2}$$
We force this equation to have the desired tangent line by substituting $x = a/(1-e)$ and determining the condition under which the resulting quadratic in $y$ has a single solution. Substitution gives this quadratic
$$\frac{a^2}{(1-e)^2} + y^2 = \left(\;p + \frac{ea\cos\alpha}{1-e}+ey\sin\alpha \;\right)^2 \tag{3}$$
whose discriminant is
$$ 4 (1 - e)^3 \left(\;p^2 (1- e) + 2 p a e \cos\alpha - a^2 ( 1 + e ) \;\right) \tag{4}$$
Now, $(3)$ will have a single solution when (and only when) $(4)$ vanishes. Solving $(4)$ for $p$ gives

$$p = a\;\frac{- e \cos\alpha \pm \sqrt{ 1 - e^2 \sin^2\alpha}}{1 - e} \tag{5}$$

The "$\pm$" accounts for the ambiguity of whether the "other" focus lies to the right or left of the origin when $\alpha = 0$.
Here is a plot for $a = 3$ and $e = 1/2$.

