How to recover the cartesian equation of a rotated parabola from its parametric equation obtained using linear algebra ( rotation matrix)?

Question

Every point of the $y = x^2$ parabola can be seen as the endpoint of a position vector of the form $< t, t^2>$.

So, rotating this curve counterclockwise can be assimilated to applying a linear transformation to this generic position vector, that is to say, applying the matrix the columns of which are the images of the bases $(1,0)$ and $(0,1)$ under a counterclockwise rotation by $R$ radians:

$\begin{bmatrix} \cos (R) & -\sin(R)\\\sin(R) & \cos(R)\end{bmatrix} \begin{pmatrix}t\\t^2\end{pmatrix} = \begin{pmatrix}\cos(R) t- \sin(R) t^2\\\sin(R) t + \cos(R) t^2\end{pmatrix}$

My question is, how to recover the cartesian equation of the rotated curve.

That is to say, how to eliminate the parameter $t$ in these equations :

$$ x = \cos(R) t- \sin(R) t^2$$

$$ y = \sin(R) t + \cos(R) t^2$$.

Note : my question is not as to what the cartesian form of the equation looks like, but as to how to recover it using this linear algebra method ( being given that the geometric prof of the rotation formula is not much intuitive).

Answer 1

You can start by writing the general form of a conic section $$ax^2+bxy+cy^2+dx+ey+f=0$$ Then plug in your forms for $x$ and $y$. Your equations are obeyed for any $t$ if you can identify the coefficients for each power of $t$ and set them to $0$ $$a(t\cos R-t^2\sin R)^2+b(t\cos R-t^2\sin R)(t\sin R+t^2\cos R)+\\c(t\sin R+t^2\cos R)^2+d(t\cos R-t^2\sin R)+e(t\sin R+t^2\cos R)+f=0$$

You can immediately see that $f=0$. Then for the coefficient of $t$ you get $$d\cos R+e\sin R=0$$ and so on, until the coefficient of $t^4$ is given by $$a\sin^2R+b\sin R\cos R+c\cos^2R=0$$ Notice that you have four equations with five unknowns, so you fix one of the $a,b,c,d,e$ values and calculate the rest.

Answer 2

From

$$ x = \cos(R) t- \sin(R) t^2$$

$$ y = \sin(R) t + \cos(R) t^2$$

Solve for $t$ and $t^2$ in terms of $x$ and $y$, you will get

$ t = \cos(R) x + \sin(R) y $

$ t^2 = - \sin(R) x + \cos(R) y $

Hence,

$ ( \cos(R) x + \sin(R) y )^2 = (- \sin(R) x + \cos(R) y ) $

Expand

$ \cos^2(R) x^2 + \sin(R)^2 y^2 + 2 \cos(R) \sin(R) x y = - sin(R) x + \cos(R) y $

which is of the form

$ a x^2 + b x y + c y^2 + d x + e y + f = 0 $

with

$ a = \cos^2(R) , b = 2 \cos(R) \sin(R) , c = \sin^2(R), d = \sin(R) , e = - \cos(R), f = 0 $