Let a curve be given in the parametrized form by:
$r(t) = (2\cos t, 2\sin t), 0 \leq t \leq 2\pi$
Find the equations of the tangents to the curve at each of its points $(X_0, Y_0)$.
Having gone through some text, it never really directly approaches a problem such as this. I have read through some articles on this website, and it provides varying solutions to this type of question and as such I can't tell what method is correct.
From the methods I have used for tangent vectors at a certain point:
$x = X_0 - 2s\sin t$ is the tangent equation for $X_0$ where $X_0 = 2\cos t$ for some $t$
$y = Y_0 + 2s\cos t$, for $Y_0$ where $Y_0= 2\sin t$ for the same $t$
If anyone could clarify if I am approaching this the right way or provide the correct method but not the answer then that would be appreciated.
The vector that locates the loci of points to the circle is given parametrically by
$$\vec r(t)=\hat x 2\cos t+\hat y2\sin t$$
At point $t_0$, we denote the vector $\vec r_0=\vec r(t_0)$. The unit tangent $\hat u(t_0)$ to the curve at $t_0$ is
$$\begin{align} \hat u(t_0)&=\left.\frac{\frac{d\vec r(t)}{dt}}{\left|\frac{d\vec r(t)}{dt}\right|}\right|_{t=t_0}\\\\ &=-\hat x\sin t_0+\hat y\cos t_0 \end{align}$$
The parametric equation of a line tangent to the circle at $\vec r_0$ is
$$\begin{align} \vec R(s)&=\vec r_0+s\vec u_0\\\\ &=\hat x (2\cos t_0-s\sin t_0)+\hat y(2\sin t_0+s\cos t_0) \end{align}$$
$$\bbox[5px,border:2px solid #C0A000]{X(s)=2\cos (t_0)-\sin (t_0)\,s}$$
$$\bbox[5px,border:2px solid #C0A000]{Y(s)=2\sin (t_0)+\cos (t_0)\,s}$$
Eliminating $s$ gives the equation of the line as
$$\bbox[5px,border:2px solid #C0A000]{y_0Y=4-x_0X}$$