How to find cartesian coordinate of velocity of particle on the trajectory, $Ax^2 +2Bxy +Cy^2=1, A,B,C >0.$ Consider a particle with constant speed $|w|=w_o$ moving on trajectory $Ax^2 +2Bxy +Cy^2=1, A,B,C >0.$ Could anyone advise me how to express cartesian coordinates of $v$ in terms of $x$ and $y \ ?$
Here is my attempt: Let $x=u\text{cos}\theta-v\text{sin}\theta, \ y=u\text{sin}\theta+v\text{cos}\theta,$ where $\text{tan}(2\theta) = \dfrac{B}{A-C}.$ Then the equation is reduced to one in terms of only $u^2$ and $v^2.$ So the space curve $r(u)=(u,g(u))$, for some function $g.$ And $w=\dfrac{dr(u)}{dt}= (\dfrac{du}{dt}, \dfrac{dg}{du} \dfrac{du}{dt}) \ ?$
 A: $$Ax^2 + 2Bxy + Cy^2 = 1$$
$$2Axdx + 2Bydx+2Bxdy + 2Cydy = 0$$
$$2(Ax+By)dx + 2(Bx+Cy)dy = 0$$
$$\Rightarrow (Ax+By)\frac{dx}{dt} + (Bx+Cy)\frac{dy}{dt}= 0$$
$$v_x= - \frac{Bx+Cy}{Ax+By}v_y $$
And we know that
$$v_x^2+v_y^2=w_0^2$$
Using these two equations you can solve for $v_x$ and $v_y$
A: Let's rewrite equation
$$
Ax^2 + 2Bxy + Cy^2 = 1
$$
as
$$
(Ax + By)^2 + (CA - B^2)y^2 = A.
$$
We have three cases.
I. $CA - B^2 = D^2 > 0$. Equation is
$$
(Ax + By)^2 + (Dy)^2 = A.
$$
It's ellipse. Let's use standard parametrization:
$$
\left\{\begin{aligned}
Ax + By &= \sqrt A\cos\phi,\\
Dy &= \sqrt A\sin\phi
\end{aligned}\right.
\Longrightarrow
\left\{\begin{aligned}
x &= \frac{1}{\sqrt A}\left(\cos\phi - \frac{B}{D}\sin\phi\right),\\
y &= \frac{\sqrt A}{D}\sin\phi
\end{aligned}\right.
$$
So, tangent vector (“speed”) is
$$
\vec p = \left(\frac{\partial x}{\partial\phi},\, \frac{\partial y}{\partial\phi}\right) = \left(-\frac{1}{\sqrt A}\left(\sin\phi + \frac{B}{D}\cos\phi\right),\,
\frac{\sqrt A}{D}\cos\phi\right)
$$
But length of $\vec p$ doesn't equal to one. Ok, let's use $\vec n = \vec p/|\vec p|$! And velocity vector became $\vec u = w_0\vec n$. Let's find them. First of all,
$$
\vec p = \left(-\frac{1}{\sqrt A}\left(\sin\phi + \frac{B}{D}\cos\phi\right),\,
\frac{\sqrt A}{D}\cos\phi\right) = \\=
\left(-\frac{1}{A}\left(Dy + \frac{B}{D}(Ax + By)\right),\,
\frac{Ax + By}{D}\right)=\\=
\frac1D \left(-\frac{1}{A}\left(D^2y + B(Ax + By)\right),\,
Ax + By\right)=\\=
\frac1D \left(-(Bx + Cy),\, Ax + By\right)
$$
and
$$
|\vec p|^2 = \frac{1}{D^2} \big((Bx + Cy)^2 + (Ax + By)^2\big) \Longrightarrow
|\vec p| = \frac{1}{D^2} \sqrt{(Bx + Cy)^2 + (Ax + By)^2}.
$$
So,
$$
\vec u = w_0 \vec n = w_0 \frac{\left(-(Bx + Cy),\, Ax + By\right)}{\sqrt{(Bx + Cy)^2 + (Ax + By)^2}}
$$
I suppose, you can now consider cases $CA - B^2 < 0$ (with $\cosh$ and $\sinh$; it's hyperbola) and $CA - B^2 = 0$ (it's pair of lines).
