Parametric curve tangency

Taking $\frac{dy}{dt} = 2t-1$ and $\frac{dx}{dt} = 2t+1$ , the slope of $y(x)$ can be computed per $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} = \frac{2t-1}{2t+1}.$ The tangent line for all points $(x(t),y(t))$ on this curve is expressible as:

$y - (t^2-t+1) = ( \frac{2t-1}{2t+1}) \cdot (x - (t^2+t+1))$

and if this tangent line is to pass through the point $(1,1)$ , then we obtain:

$1 - (t^2-t+1) = ( \frac{2t-1}{2t+1}) \cdot (1 - (t^2+t+1))$ ;

or $t-t^2 = ( \frac{2t-1}{2t+1}) \cdot (-t^2-t)$ ;

or $1-t = - ( \frac{2t-1}{2t+1}) \cdot (1+t)$ ;

or $(t-1)(2t+1) = (2t-1)(t+1);$

or $2t^2 -t-1 = 2t^2 + t -1;$

or $-t = t;$

or $\boxed{t=0 \Rightarrow (x(0),y(0)) = (1,1)}$ . Hence, there is only $\boxed{ONE}$ such tangent line that satisfies the above initial parametric conditions.