Answer
$$y=x+2$$
Work Step by Step
Recall, to find the equation of a line, we need to find the slope of the line and a point that lies on the line.
Now, we want to find the equation of the line tangent to the curve at the point where $t=1$.
So, we know the slope of this line is the derivative of the curve at the point where $t=1$, and because this line is tangent to the curve at the point where $t=1$, we know the point where $t=1$ lies on this line.
So, let us first find the derivative at $t=1$. The curve is given by the parametric equations $$x=t-t^{-1}$$ and $$y=1+t^2.$$ So, the derivative at $t=1$ is $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}},$$ evaluated at $t=1$
So, we find the derivatives of each of the parametric equations with respect to $t$. $$\frac{dx}{dt}=\frac{d}{dt}(t)-\frac{d}{dt}(t^{-1})=1--(t)^{-2}=1+t^{-2}$$ and $$\frac{dy}{dt}=\frac{d}{dt}(1)+\frac{d}{dt}(t^2)=0+2t^1 =2t.$$ Then $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t}{1+t^{-2}}.$$ Next we plug in $t=1$ to get $$\frac{2(1)}{1+1^{-2}}=\frac{2}{1+\frac{1}{1^2}}=\frac{2}{1+\frac{1}{1}}=\frac{2}{1+1}=\frac{2}{2}=1$$ Thus, the slope of our line is $1$.
Now, to find the point that lies on the line, we plug $t=1$ into our parametric equations. We get $$x=t-t^{-1}=1-1^{-1}=1-1=0$$ and $$y=1+t^2=1+1^2=1+1=2.$$ Thus, our point is $(0,2)$.
Finally, we are ready to find the equation of our line. We will use the formula $$y-y_{1}=m(x-x_{1}),$$ where $m$ is the slope and $(x_{1},y_{1})$ is the point that lies on the line.
For us, $m=1$ and $(x_{1},y_{1})=(0,2)$ We plug these into the equation to get $$y-2=1(x-0).$$ Thus, we have $$y-2=x.$$ Adding $2$ to both sides, we get $$y=x+2.$$
