Can ellipse equation be transformed through one of its foci? Can we transform ellipse equation to represent an ellipse transformed by tilting it through its focus such that its center point moves in circular manner and one of its focus stays at constant position? If this is possible, how can we achieve this kind of equation?
 A: Suppose $f:=\begin{bmatrix}f_1\\f_2\end{bmatrix}$ is one of the foci. And $\theta$ is the angle you want to rotate.
For every point $x:=\begin{bmatrix}x_1\\x_2\end{bmatrix}$ of the ellipse do this. 


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*$x-f=\begin{bmatrix}x_1-f_1\\x_2-f_2\end{bmatrix}$ This brings the focus $f$ to the origin.

*$A(x-f)=\begin{bmatrix}\cos(\theta)(x_1-f_1)+\sin(\theta)(x_2-f_2)\\-\sin(\theta)(x_1-f_1)+\cos(\theta)(x_2-f_2)\end{bmatrix}$, where $A=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{bmatrix}$. This rotates the ellipse around the origin (which is where the focus is now.)

*$A(x-f)+f=\begin{bmatrix}\cos(\theta)(x_1-f_1)+\sin(\theta)(x_2-f_2)+f_1\\-\sin(\theta)(x_1-f_1)+\cos(\theta)(x_2-f_2)+f_2\end{bmatrix}$. This returns the focus $f$ to where it was before such that the ellipse looks rotated around its focus $f$.
So, if your original equation is $q(x_1,x_2)=0$, just replace $x_1,x_2$ by $\cos(\theta)(x_1-f_1)+\sin(\theta)(x_2-f_2)+f_1$ and $-\sin(\theta)(x_1-f_1)+\cos(\theta)(x_2-f_2)+f_2$ respectively.
A: Let the angle rotated be $z$ about the fixed focus
and lets take the standard equation of ellipse
Use the parametric form of line passing through the fixed focus and with slope $\tan z$ to find the point through which the nearer directrix passes
like,   $(x-ae)/\cos z = (y-0)/ \sin z = -r$ ,where $r= a/e -ae$
find $x$ and $y$ and hence the equation of directrix with slope $-\cot z$
now you know the focus and directrix so find the equation of ellipse.
