Find affine transformations that map the canonical curves…

Find affine transformations that map the canonical curves $x'^2+y'^2=1$, $x'^2-y'^2=1$ and $y'^2=x'$ onto an ellipse, a hyperbola and a parabola with given centre (vertex), semi-axes (parameter) and axis direction $\theta$.

Step 1: Ellipse

An ellipse with centre $(x_0,y_0)$, semi-axes $a,b$ and major axis at angle $\theta$ has the parametrization $\vec r=\vec r_0+a\cos t\,\vec u+b\sin t\,\vec v$, with $\vec u=(\cos\theta,\sin\theta)$ and $\vec v=(-\sin\theta,\cos\theta)$. Putting $x'=\cos t,\ y'=\sin t$:

\[ \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}a&0\\0&b\end{pmatrix}\begin{pmatrix}x'\\y'\end{pmatrix}+\begin{pmatrix}x_0\\y_0\end{pmatrix}. \]

\[x=a x'\cos\theta-b y'\sin\theta+x_0,\qquad y=a x'\sin\theta+b y'\cos\theta+y_0\]

Step 2: Hyperbola

The same map applied to $x'^2-y'^2=1$ gives the hyperbola with centre $(x_0,y_0)$, real semi-axis $a$ along $\vec u$ and imaginary semi-axis $b$ along $\vec v$: $x=a x'\cos\theta-b y'\sin\theta+x_0,\qquad y=a x'\sin\theta+b y'\cos\theta+y_0$.

Check with $\theta=0$: $x'=\tfrac{x-x_0}{a}$, $y'=\tfrac{y-y_0}{b}$, so $\tfrac{(x-x_0)^2}{a^2}-\tfrac{(y-y_0)^2}{b^2}=1$.

Step 3: Parabola

For $y^2=2px$, put $x=x'$, $y=\sqrt{2p}\,y'$; then $y'^2=x'$ becomes $y^2=2px$. For a vertex $(x_0,y_0)$ and axis direction $\theta$: \[x=x'\cos\theta-\sqrt{2p}\,y'\sin\theta+x_0,\qquad y=x'\sin\theta+\sqrt{2p}\,y'\cos\theta+y_0\]

No imaginary factors are needed. The earlier remark about “$i/\sqrt{|p|}$” is unnecessary: the sign of $p$ is handled by $\theta$ (turning by $\pi$ reverses the opening).