Given the curve y=(x²)/(x²-1), write the equation

Given the curve $y=\dfrac{x^{2}}{x^{2}-1}$, write the equation of the tangent at an arbitrary point, then find the coordinates of the point at which this tangent meets the curve again.

Step 1: Derivative

\[ y' = \frac{2x(x^{2}-1)-x^{2}\cdot2x}{(x^{2}-1)^{2}} = \frac{-2x}{(x^{2}-1)^{2}}. \tag{1} \]

Step 2: Tangent at an arbitrary point

Let $P=(t,\,t^{2}/(t^{2}-1))$, $t\neq\pm1$, with slope $k=-2t/(t^{2}-1)^{2}$. Expanding the point–slope form and combining $2t^{2}+t^{2}(t^{2}-1)=t^{2}(t^{2}+1)$, \[\text{Tangent: }\quad y = \frac{-2t\,x+t^{2}(t^{2}+1)}{(t^{2}-1)^{2}}. \tag{2}\]

Step 3: Intersect the tangent with the curve

Substituting $y=x^{2}/(x^{2}-1)$ into (2), cross-multiplying, and collecting by powers of $x$ (the $x^2$ coefficient simplifies via $(t^2-1)^2-t^2(t^2+1)=1-3t^2$): \[2t\,x^{3} + (1-3t^{2})x^{2} - 2t\,x + t^{2}(t^{2}+1) = 0. \tag{3}\]

Step 4: Factor out the double root $x=t$

Dividing twice by $(x-t)$ (each step checked by substituting $x=t$): \[2t\,x^{3}+(1-3t^{2})x^{2}-2t\,x+t^{2}(t^{2}+1) = (x-t)^{2}\bigl(2tx+t^{2}+1\bigr). \tag{4}\]

Step 5: The second intersection point $B$

The third root comes from $2tx+t^{2}+1=0$: \[x_{B} = -\frac{t^{2}+1}{2t}. \tag{5}\]

Then $x_{B}^{2}=\dfrac{(t^{2}+1)^{2}}{4t^{2}}$ and $x_{B}^{2}-1=\dfrac{(t^{2}-1)^{2}}{4t^{2}}$, so

\[ y_{B} = \frac{x_{B}^{2}}{x_{B}^{2}-1} = \frac{(t^{2}+1)^{2}}{(t^{2}-1)^{2}}. \tag{6} \]

Step 6: Verification

\[B \text{ on the curve: } \frac{x_B^2}{x_B^2-1}=\frac{(t^2+1)^2/4t^2}{(t^2-1)^2/4t^2}=\frac{(t^2+1)^2}{(t^2-1)^2}=y_B\ \checkmark\]

\[B \text{ on the tangent (2): } \frac{-2t\!\left(-\tfrac{t^2+1}{2t}\right)+t^2(t^2+1)}{(t^2-1)^2}=\frac{(t^2+1)(1+t^2)}{(t^2-1)^2}=y_B\ \checkmark\]