Tangent line to parametric curve at a point Consider the curve given parametrically by $(x,y,z)=(2−t,−1−t^2,−2t−3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(2,8,−162)$. Find the coordinates of the point $P$. 
Can anyone give me a hand with this one? I honestly have no idea where to start other than rewriting the curve as, 
$\vec{r(t)}$ = $\langle 2-t, -1-t^2, -2t-3t^3\rangle$, and then taking the derivate to obtain,
$\vec{r'(t)}$ = $\langle -1, -2t, -2 -9t^2\rangle$
 A: Let $\vec{r}(t) = \langle 2−t,−1−t^2,−2t−3t^3\rangle$, where $t\in \mathbb{R}$. Then $\vec{r}'(t) = \langle -1, -2t, -2 -9t^2\rangle$. 
For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by 
$$
\vec{T}(t) = t \:\vec{r}'(t_0) + \vec{r}(t_0), \mbox{ where } t\in \mathbb{R}. 
$$
Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$. 
Thus, setting the following equal to each other, 
\begin{align*}
t\langle -1, -2t, -2 -9t^2\rangle + \langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3\rangle = \langle 2,8, -162 \rangle, 
\end{align*}
we need to solve for $t_0$. 
The above vector equation gives us three equations: 
\begin{align*}
-t+2-t_0&=2,  \hspace{4mm} (\dagger) \\ 
-2t_0t-1-t_0^2&=8,  \hspace{4mm} (\ddagger)  \\ 
-2t-9t_0^2t-2t_0-3t_0^3&=-162.  \hspace{4mm} (\Omega)\\ 
\end{align*}
Use the first equation $(\dagger)$ to solve for $t:$ 
$$
t=-t_0. 
$$
Let's substitute this into the second equation $(\ddagger)$: 
$$
2t_0^2 -1-t_0^2 = 8, 
$$
which gives us $t_0^2 = 9$. 
So $t_0 = \pm 3$ while $t=\mp 3$. 
If $t_0=3$, then $t=-3$. We substitute these into the third equation $(\Omega)$ to obtain: 
$$
6+9\cdot 9\cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) \not= -162. 
$$
We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(\Omega)$: 
$$
-6-9\cdot 9\cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162. 
$$ 
So $P$ has position vector $\vec{r}(-3)=\langle 5,-10,87 \rangle $, which is when its tangent line passes through the point $(2,8,−162)$.
A: Once $\vec{r'}=(-1,-2t,-2-9t^2)$ the tangent line is given by $$\vec t=\lambda\vec{r'}\text{, }\lambda \in\ \Bbb R$$
So, we can write
$$\vec P+\vec t=(2,8,-162)\to\\
(2-t,-1-t^2,-2t-3t^3)+(-\lambda,-2\lambda t,-2\lambda-9\lambda t^2)=(2,8,-162)$$
which means:
$$2-t-\lambda=2\to t=-\lambda$$
$$-1-t^2-2\lambda t=8$$
$$-2t-3t^3-9\lambda t^2=-162$$
Can you finish?
