Find all asymptotes of x=(t³)/(1-t²), y=(1+t²)/(1-t²)

Find all asymptotes of $x=\dfrac{t^3}{1-t^2}$, $y=\dfrac{1+t^2}{1-t^2}$.

Step 1: Singular parameter values

The denominator $1-t^2$ vanishes at $t=\pm1$; each sends $x$ and $y$ to infinity and must be treated separately.

Step 2: Asymptote as $t\to1$

\[ k=\lim_{t\to1}\frac{y}{x}=\lim_{t\to1}\frac{1+t^2}{t^3}=2. \tag{1} \]

For the intercept, form $y-2x$ and factor, using $t=1$ as a common root of numerator and denominator: \[y-2x=\frac{1+t^2-2t^3}{1-t^2}=\frac{-(t-1)(2t^2+t+1)}{-(t-1)(t+1)}=\frac{2t^2+t+1}{t+1}. \tag{2}\]

\[ b=\lim_{t\to1}\frac{2t^2+t+1}{t+1}=\frac{4}{2}=2. \tag{3} \]

Asymptote: $y=2x+2$.

Step 3: Asymptote as $t\to-1$

\[ k=\lim_{t\to-1}\frac{y}{x}=\lim_{t\to-1}\frac{1+t^2}{t^3}=-2. \tag{4} \]

\[ y+2x=\frac{1+t^2+2t^3}{1-t^2}=\frac{(t+1)(2t^2-t+1)}{-(t-1)(t+1)}=\frac{2t^2-t+1}{1-t}. \tag{5} \]

\[ b=\lim_{t\to-1}\frac{2t^2-t+1}{1-t}=\frac{4}{2}=2. \tag{6} \]

Asymptote: $y=-2x+2$.

Step 4: Verification

\[t=0.999\text{: }x\approx499.25,\ y\approx999.50,\ y-2x-2\approx0.0005\to0\ \checkmark\]

\[t=-0.999\text{: }x\approx-499.25,\ y\approx999.50,\ y+2x-2\approx0.0005\to0\ \checkmark\]