Answer
Tangent line: $\quad y=-x-1$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=1}=1$
Work Step by Step
The slope of the tangent line at $t=1$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=1}$
$\displaystyle \left\{\begin{array}{l}
\frac{dx}{dt}=-t^{-2},\\
\\
\frac{dy}{dt}=t^{-1}
\end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{t^{-1}}{-t^{-2}}=-t$
$\left. \displaystyle \frac{dy}{dx} \right|_{t=1}=-1$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-1}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[-t]=-1$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{-1}{-t^{-2}}=t^{2}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=1}=1$
The point $P$ on the curve corresponding to $t=1$ is
$\left\{\begin{array}{l}
x=1\\
y=-2
\end{array}\right.\quad, P(1,-2)$
Use the point-slope equation for the tangent line.
$y+2=-1\cdot(x-1)$
$y=-x+1-2$
$y=-x-1$
