Bounds of rotated egg? I'm attempting to find the bounding box for a rotated egg shape without resorting to interpolation if at all possible.  In researching this I've come across this link which lists useful equations for plotting the $(x, y)$ coordinates of an egg shape as:
\begin{align*}
  x &= \frac{r((c-2) \cos \theta + c + 2)(\cos \theta + 1)}{4}, \\
  y &= r \cdot \sin \theta.
\end{align*}
But if I were to normalize it to the centerpoint of the circular area—$(0, 1)$ in the image below, then rotate it by some arbitrary angle about that new centerpoint­—how can the $x$ and $y$ bounds be found without interpolation?

 A: Instead of rotating your egg I'd prefer to rotate the bounding box. I hope you are able to translate between these views. Suppose $(a,b)$ is one direction of the box. Then in order to find the points where the curve is maximal or minimal with respect to that direction, you need to find the points where the tangent of the curve is orthogonal to that direction. Orthogonal means dot product is zero, and the tangent direction is the direction of derivatives. So let's start by computing some derivatives (for which I used Sage):
\begin{align*}
\frac{\mathrm d\,x}{\mathrm d\theta}&=
\frac12r\sin\theta\bigl((2-c)\cos\theta-c\bigr) &
\frac{\mathrm d\,y}{\mathrm d\theta}&=
r\cos\theta
\end{align*}
Now set the dot product equal to zero:
\begin{align*}
0&\overset!=\left\langle\begin{pmatrix}a\\b\end{pmatrix},
\begin{pmatrix}\frac{\mathrm d\,x}{\mathrm d\theta}\\
\frac{\mathrm d\,y}{\mathrm d\theta}\end{pmatrix}\right\rangle=
\frac12ar\sin\theta\bigl((2-c)\cos\theta-c\bigr)+br\cos\theta\\
0&\overset!=a\sin\theta\bigl((2-c)\cos\theta-c\bigr)+2b\cos\theta
\end{align*}
Now dealing with trigonometric functions here may make things a bit hard, so I'd use the tangent half-angle substitution here:
$$t:=\tan\frac\theta2\qquad
\sin\theta=\frac{2t}{1+t^2}\qquad
\cos\theta=\frac{1-t^2}{1+t^2}$$
Using this you can write
\begin{align*}
0&\overset!=a\frac{2t}{1+t^2}\bigl((2-c)\frac{1-t^2}{1+t^2}-c\bigr)+2b\frac{1-t^2}{1+t^2}\\
0&\overset!=at\bigl((2-c)(1-t^2)-c(1+t^2)\bigr)+b(1-t^2)(1+t^2)\\
0&\overset!=bt^4+2at^3+2a(c-1)t-b
\end{align*}
This is a quartic (i.e. 4th degree) polynomial equation in $t$. Solving those is not beautiful, but you can solve this using radical expressions, without interpolation or resorting to numerical methods. I would expect two real solutions, one corresponding to each boundary. Turn the values $t$ you get back into $\theta$ (or compute $\sin\theta$ and $\cos\theta$ directly from $t$ to avoid the trigonometry) then plug this back into the original formula to obtain the extremal points. Do the same in the orthogonal direction $(-b,a)$ and you have all four points defining your bounding box. Rotate that if you want it aligned with your axes.
Note that the tangent half-angle substitution has a blind spot: $\theta=\pm\pi$ corresponds to $t\to\infty$. So if your direction is not rotated, and you are looking for vertical tangents, be sure to cover that case separately.
Below is a snapshot of one instance of this problem.
See this blog page for an interactive version with some additional text.

