A tangent is drawn at an arbitrary point of the curve y =

A tangent is drawn at an arbitrary point of the curve $y = x+\dfrac{1}{x^{2}}$. Determine the coordinates of the point $M$ at which this tangent meets the curve again.

Step 1: Derivative and tangent at an arbitrary point

Let $P=\left(t,\,t+\dfrac1{t^{2}}\right)$, $t\neq0$. Since $y'=1-\dfrac2{x^{3}}$, the slope is $k=1-\dfrac2{t^{3}}$, and expanding the point–slope form (the constants $-t+\tfrac2{t^2}+t+\tfrac1{t^2}$ collapse to $\tfrac3{t^2}$):

\[\text{Tangent: }\quad y = \left(1-\frac{2}{t^{3}}\right)x + \frac{3}{t^{2}}. \tag{1}\]

Step 2: Intersect the tangent with the curve

Substituting $y=x+1/x^{2}$ into (1) and multiplying through by $x^{2}t^{3}$: \[2x^{3} - 3t\,x^{2} + t^{3} = 0. \tag{2}\]

Step 3: Factor out the double root $x=t$

Dividing twice by $(x-t)$, first $2x^3-3tx^2+t^3=(x-t)(2x^2-tx-t^2)$, then $2x^2-tx-t^2=(x-t)(2x+t)$, so

\[ 2x^{3}-3tx^{2}+t^{3} = (x-t)^{2}(2x+t). \tag{3} \]

Step 4: The second intersection point $M$

The third root is $x_{M}=-t/2$, and

\[ y_{M} = x_{M}+\frac{1}{x_{M}^{2}} = -\frac{t}{2}+\frac{4}{t^{2}}. \tag{4} \]

Step 5: Verification

\[M \text{ on the curve: } x_M+1/x_M^2=-t/2+4/t^2=y_M\ \checkmark\]

\[M \text{ on the tangent (1): } \left(1-\tfrac{2}{t^3}\right)\!\left(-\tfrac{t}{2}\right)+\tfrac{3}{t^2}=-\tfrac{t}{2}+\tfrac{1}{t^2}+\tfrac{3}{t^2}=-\tfrac{t}{2}+\tfrac{4}{t^2}=y_M\ \checkmark\]

\[\text{Double root at } x=t\text{: } (x-t)^2(2x+t)\big|_{x=t}=0\ \checkmark\]