How do I get a tangent to a rotated ellipse in a given point?

I have just graduated from a school you would call High School and even though we talked about tangents to ellipses, we never covered rotated ellipses. So, what I am looking for, is a formula for a tangent to a rotated ellipse. I had searched the internet for solutions, but unfortunately did not come across any solutions. I hope you can help me.

What I have is an ellipse:

$(X_c,Y_c)$ = center of the ellipse.

$\phi$ = angle between the $X$-axis and the major axis of the ellipse.

$t \in [0,2\pi[$

$$\begin{cases}x = X_c + a \cos(t) \cos(\phi) - b \sin(t) \sin(\phi) \\ y = Y_c + a \cos(t) \sin(\phi) + b \sin(t) \cos(\phi) \end{cases}$$

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no, but the rotation does... –  the_critic Jul 17 '12 at 14:11
Do you know about derivatives? –  Raskolnikov Jul 17 '12 at 14:12
The tangent at the point $((x(t_0),y(t_0))$ will be given (parametrically) by the line $((x(t_0),y(t_0)) + \lambda (\frac{d x(t_0)}{dt}, \frac{d y(t_0)}{dt})$, where $\lambda \in \mathbb{R}$. –  copper.hat Jul 17 '12 at 14:44

The slope of the line will be $\dfrac {b\cos(t)\cos(\phi)-a\sin(t)\sin(\phi)}{-b\cos(t)\cos(\phi)-a\sin(t)\sin(\phi)}$

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First of all, thank you. Are you sure this is right ? Unfortunately it gives me wrong results. –  the_critic Jul 17 '12 at 16:23
Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$. –  André Nicolas Jul 17 '12 at 18:10
im sorry what i first posted is wrong as andre nicolas pointed out. that was the rate of change with respect to t. The new formula should be right. sorry for the inconvenience –  dREaM Jul 17 '12 at 18:44