Answer
$t=-2\to (x,y)=(\ln 5,-1)$
$t=-1\to (x,y)=(\ln 2,-\frac{1}{3})$
$t=0\to (x,y)=(0,0)$
$t=1\to (x,y)=(\ln 2,\frac{1}{5})$
$t=2\to (x,y)=(\ln 5,\frac{1}{3})$
Work Step by Step
Given: $x=\ln(t^2+1)$ and $y=\frac{t}{t+4}$
For $t=-2$,
$x=\ln ((-2)^2+1)=\ln 5$
$y=\frac{-2}{-2+4}=\frac{-2}{2}=-1$
So, $(x,y)=(\ln 5,-1)$ at $t=-2$
For $t=-1$,
$x=\ln ((-1)^2+1)=\ln 2$
$y=\frac{-1}{-1+4}=\frac{-1}{3}$
So, $(x,y)=(\ln 2,-\frac{1}{3})$ at $t=-1$
For $t=0$,
$x=\ln (0^2+1)=\ln 1=0$
$y=\frac{0}{0+4}=\frac{0}{4}=0$
So, $(x,y)=(0,0)$ at $t=0$
For $t=1$,
$x=\ln (1^2+1)=\ln 2$
$y=\frac{1}{1+4}=\frac{1}{5}$
So, $(x,y)=(\ln 2,\frac{1}{5})$ at $t=1$
For $t=2$,
$x=\ln (2^2+1)=\ln 5$
$y=\frac{2}{2+4}=\frac{2}{6}=\frac{1}{3}$
So, $(x,y)=(\ln 5,\frac{1}{3})$ at $t=2$
