Showing that $(x,y)=\left(\frac{c\sin t+d}{a+b\cos t},\frac{e\cos t+f}{a+b\cos t}\right)$ parameterizes an ellipse I want to show that
$$\mathbf{P}(t) = (X, Y) =  \left( \dfrac{ c \sin t + d}{a + b \cos t} , \dfrac{ e \cos t + f }{a + b \cos t } \right)$$
is actually an ellipse, given that $ | b | \lt | a | $
How can I prove this?
What I have tried:
Guided by @orangeskid solution to a recent problem, I've defined the vector
$ \mathbf{r}(t) = (x(t),y(t), z(t)) = (c \sin t + d , e \cos t + f , a + b \cos t ) $
which is an ellipse in $3D$ space whose parametric equation is
$ \mathbf{r}(t) = [d, f, a]^T + \cos t \ [ 0, e, b ]^T + \sin t \ [c , 0, 0 ]^T $
Using matrix-vector notation, this can be written as
$ \mathbf{r}(t) = M \mathbf{v}(t) \hspace{50pt}(1) $
where
$ M = \begin{bmatrix} 0 && c && d \\ e && 0 && f \\ b && 0 && a \end{bmatrix} $
and
$ \mathbf{v}(t) = \begin{bmatrix} \cos t \\ \sin t \\ 1 \end{bmatrix} $
Now, we note that
$ \mathbf{v}^T(t) Q_0 \mathbf{v}(t) = 0 \hspace{50pt}(2) $
where
$Q_0 = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix} $
From eq. $(1)$,
$ \mathbf{v}(t) = M^{-1} \mathbf{r}(t) \hspace{50pt} $
Substituting this into $(2)$,
$\mathbf{r}^T (t) Q \mathbf{r}(t) = 0 \hspace{50pt}(3) $
where
$ Q = M^{-T} Q_0 M^{-1} $
Thus, $(3)$ becomes
$\begin{bmatrix} x && y && z \end{bmatrix} \begin{bmatrix} Q_{11} && Q_{12} && Q_{13} \\ Q_{12} && Q_{22} && Q_{23} \\ Q_{13} && Q_{23} && Q_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 $
Dividing both sides of the above equation by $z^2$, we get
$\begin{bmatrix} \dfrac{x}{z} && \dfrac{y}{z} && 1 \end{bmatrix} \begin{bmatrix} Q_{11} && Q_{12} && Q_{13} \\ Q_{12} && Q_{22} && Q_{23} \\ Q_{13} && Q_{23} && Q_{33} \end{bmatrix} \begin{bmatrix} \dfrac{x}{z} \\ \dfrac{y}{z} \\ 1 \end{bmatrix} = 0 $
But, $ \dfrac{x}{z} = P_x = X, \dfrac{y}{z} = P_y = Y $, hence,
$\begin{bmatrix} X && Y && 1 \end{bmatrix} \begin{bmatrix} Q_{11} && Q_{12} && Q_{13} \\ Q_{12} && Q_{22} && Q_{23} \\ Q_{13} && Q_{23} && Q_{33} \end{bmatrix} \begin{bmatrix} X \\ Y \\ 1 \end{bmatrix} = 0 $
And this an equation of a conic in $X$ and $Y$.  Further investigation is needed to determine if it is indeed an ellipse.

Try to solve $\cos t$ and $\sin t$ in terms of $x$ and $y$:
\begin{align}
  \cos t &= \frac{f-ay}{by-e} \\
  \sin t &= \frac{(ae-bf)x+bdy-de}{c(e-by)} \\
  \cos^2 t+\sin^2 t &= 
  \left(
    \frac{f-ay}{by-e}
  \right)^2+
  \left[
    \frac{(ae-bf)x+bdy-de}{c(e-by)}
  \right]^2 \\
  c^2(by-e)^2 &=c^2(f-ay)^2+[(ae-bf)x+bdy-de]^2 \\
\end{align}
Now,
$0=
\begin{pmatrix}
  x & y & 1
\end{pmatrix}
\begin{pmatrix}
  (ae-bf)^2 & (ae-bf)bd & (bf-ae)de \\
  (ae-bf)bd & c^2(a^2-b^2)+b^2d^2 & c^2(be-af)-bd^2e \\
  (bf-ae)de & c^2(be-af)-bd^2e & c^2(f^2-e^2)+d^2e^2
\end{pmatrix}
\begin{pmatrix}
  x \\ y \\ 1
\end{pmatrix}$
and discriminants
$$\delta = (ae-bf)^2 (a^2-b^2)c^2 > 0$$
$$\Delta = -c^4 (ae-bf)^4 < 0$$
which gives a real ellipse providing $ae\ne bf$ and $c\ne 0$.
