If one knows the gradient of an unknown curve at an angle how can the unknown curve be found? I have an equation for the gradient of a curve at a given theta
$$\frac{dy(θ)}{dx(θ)}=\frac{−cos(θ)+A_x}{sin(θ)−A_y}$$
I know from geometric methods that the solution to this is a rotated ellipse, given in parametric form by:
$$x(t)=\frac{cos(t)-\sqrt{1-A_x^2-A_y^2}\frac{A_y}{A_x}\cdot sin(t)}{2\sqrt{1+{\left(\frac{A_y}{A_x}\right)}^2}}+\frac{A_x}{2}\\
y(t)=\frac{cos(t)\frac{A_y}{A_x}+\sqrt{1-A_x^2-A_y^2}\cdot sin(t)}{2\sqrt{1+{\left(\frac{A_y}{A_x}\right)}^2}}+\frac{A_y}{2}
$$
However, I can't figure out how to find this solution using calculus. 
My current approach revolves around the idea that
$$\frac{dy(\theta)}{dx(\theta)} = \frac{dy(\theta)}{d\theta} \cdot \frac{1}{\frac{dx(\theta)}{d\theta}}$$
But so far I haven't had any success using this.
Can anyone help?
Edit: 
In case i'm being unclear here is a link to a geogebra applet which shows how moving the point $(A_x,A_y)$ changes the equation of the ellipse https://www.geogebra.org/classic/k74bvwyd

I am afraid that the present wording of the question might be not consistent and/or uncomplete. So the answer below is limited to solving the ODE in polar coordinates $(\rho,\theta)$ with center $(a,b)$ :
$$\begin{cases}
x=a+\rho\cos(\theta) \\
y=b+\rho\sin(\theta)
\end{cases}$$
$$\frac{dy(θ)}{dx(θ)}=\frac{−\cos(θ)+A_x}{\sin(θ)−A_y}$$
$$\begin{cases}
dx=\cos(\theta)d\rho-\rho\sin(\theta)d\theta \\
dy=\sin(\theta)d\rho+\rho\cos(\theta)d\theta 
\end{cases}$$
$$\frac{dy}{dx}=\frac{\sin(\theta)\frac{d\rho}{d\theta}+\rho\cos(\theta) }{\cos(\theta)\frac{d\rho}{d\theta}-\rho\sin(\theta) }=\frac{−\cos(θ)+A_x}{\sin(θ)−A_y}$$
$$\big(\sin(\theta)\frac{d\rho}{d\theta}+\rho\cos(\theta)\big)(\sin(θ)−A_y)-\big(\cos(\theta)\frac{d\rho}{d\theta}-\rho\sin(\theta) \big)(−\cos(θ)+A_x) =0$$
$$\left(1-A_y\sin(\theta) -A_x\cos(\theta)\right)\frac{d\rho}{d\theta}
=A_y\rho\cos(\theta) -A_x\rho\sin(\theta) $$
$$\frac{d\rho}{\rho}=\frac{A_y\cos(\theta) -A_x\sin(\theta)}{1-A_y\sin(\theta) -A_x\cos(\theta)}d\theta $$
$$\ln(\rho)=-\ln|1-A_y\sin(\theta) -A_x\cos(\theta)|+\text{constant}$$
$$\rho=\frac{C}{1-A_y\sin(\theta) -A_x\cos(\theta)}$$
$$\boxed{\begin{cases}x(\theta)=a+\frac{C\:\cos(\theta)}{1-A_y\sin(\theta) -A_x\cos(\theta)}\\
y(\theta)=b+\frac{C\:\sin(\theta)}{1-A_y\sin(\theta) -A_x\cos(\theta)}\end{cases}}$$
In order to relate the above result $x(\theta),y(\theta)$ to the expected result $x(t),y(t)$ some relationships are lacking, but one cannot say for sure what they are. Moreover the above $x(\theta),y(\theta)$ would be different if there was a typo of sign in $\frac{−\cos(θ)+A_x}{\sin(θ)−A_y}$ . 
