Tangent to a curve with a given parametric equation We have to find all the lines that pass through the point $(1, 1)$ and are tangent to the curve represented parametrically as
$$x = 2t – t^2 \quad\mbox{and}\qquad y = t + t^2$$
I find the slope
$$y' = \dfrac{1+2t}{2-2t}$$
Now when $t=1$ the tangent is $x=1$.
But how to find when is not equal to $1$?
 A: HINT:
The equation of the tangent is $$\dfrac{y-(t+t^2)}{x-(2t-t^2)}=\dfrac{1+2t}{2-2t}$$
Now as it passes through $(1,1)$
$$\dfrac{1-(t+t^2)}{1-(2t-t^2)}=\dfrac{1+2t}{2-2t}$$
$$\iff2(1-t)(1-t-t^2)=(1+2t)(1-t)^2$$
A: If you wish for a line to pass through a point $P$ and be tangent to a curve $\gamma$ (given parametrically) then you want $\gamma'(t) = \lambda \vec{v}$ where $\vec{v}$ is the direction of the line passing through $P$. For you problem, you can:
$$\gamma(t) = \langle 2t-t^2,t+t^2 \rangle \  \ \textrm{and} \ \ \vec{v} = \langle v_1, v_2 \rangle$$
and solve the resulting system. 
A: Note: I will do this for the coordinates $(0,6)$ and $t=2$ as an example.
Use the fact that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. 
We evaluate this to be $\frac{dy}{dx}=\frac{1+2t}{2-2t}$, as you have correctly mentioned.
Now evaluate $\frac{dy}{dx}$ at the point you consider by using the value of $t$. This value at the point $(0,6)$ is $\frac{dy}{dx}=-\frac{5}{2}$.
Substitute this on $y=\frac{dy}{dx}\cdot x+c$, where $c$ is a constant to evaluate in order to align your tangent to just 'barely touch' your parametric equation.
Evaluate $c$ by substituting the value of $y=6$ and the value of $x=0$ on $y=-\frac{5}{2} x+c$ and evaluate the value of $c$. We evaluate this to be $6=-\frac{5}{2} \cdot 0 + c$. Hence, $c=6$.
Now just substitute this to obtain your tangent $y=-\frac{5}{2} x + 6$.
