Determine the line that is simultaneously a tangent
Determine the line that is simultaneously a tangent at one point and a normal at another point of the curve $x=1+\dfrac{1}{t^{2}},\ y=t+\dfrac{1}{t^{2}}$.
Explanation of Method
The same idea as the previous problem, on a curve where the answer is no longer a tidy radical.
Step 1: Slope along the curve
\[ \dot x = -\frac{2}{t^{3}}, \qquad \dot y = 1-\frac{2}{t^{3}}, \qquad \frac{dy}{dx} = \frac{t^{3}-2}{-2} = \frac{2-t^{3}}{2}. \tag{1} \]
Step 2: Where the tangent at $t_{0}$ meets the curve again
Requiring the point at $t_{1}$ to lie on the tangent at $t_{0}$, and simplifying (the terms $-2(t_1+t_0)$ cancel from both sides), the condition reduces to a quadratic in $t_{1}$:
\[ 2t_{1}^{2} - t_{0}t_{1} - t_{0}^{2} = 0 \;\Longrightarrow\; (2t_{1}+t_{0})(t_{1}-t_{0}) = 0. \tag{2} \]
Besides $t_{1}=t_{0}$, the tangent at $t_{0}$ meets the curve again at
\[ t_{1} = -\frac{t_{0}}{2}. \tag{3} \]
Step 3: Perpendicularity condition
The same line is the normal at $t_{1}$ precisely when the two tangent slopes are negative reciprocals: \[\frac{2-t_{0}^{3}}{2}\cdot\frac{2-t_{1}^{3}}{2} = -1 \;\Longrightarrow\; (2-t_{0}^{3})(2-t_{1}^{3}) = -4. \tag{4}\]
Substitute $t_{1}=-t_{0}/2$, so $t_{1}^{3}=-t_{0}^{3}/8$; with $u=t_{0}^{3}$, \[(u-2)\!\left(-\frac{u}{8}-2\right) = -4 \;\Longrightarrow\; (u-2)(u+16) = 32 \;\Longrightarrow\; u^{2}+14u-64=0. \tag{5}\]
Step 4: Solve for $u=t_{0}^{3}$
\[ u = \frac{-14\pm\sqrt{196+256}}{2} = \frac{-14\pm\sqrt{452}}{2} = -7\pm\sqrt{113}. \tag{6} \]
Both values are real, so both give a real cube root $t_{0}=\sqrt[3]{u}$; numerically $u_{1}=-7+\sqrt{113}\approx3.630$ and $u_{2}=-7-\sqrt{113}\approx-17.630$, giving $t_{0}\approx1.5369$ and $t_{0}\approx-2.6027$ respectively.
Step 5: The line itself
Writing the tangent at $t_{0}$ in slope–intercept form (using $y_0=t_0+1/t_0^2$, $x_0=1+1/t_0^2$, and simplifying the intercept with $1-k_0=u/2$): $y = \frac{2-t_{0}^{3}}{2}\,x + \frac{t_{0}^{3}+3t_{0}-2}{2}, \qquad t_{0}^{3} = -7\pm\sqrt{113}$.
There are two such lines, one for each real root of (5).
Step 6: Verification
\[t_0\approx1.53688\text{: } (2-t_0^3)(2-t_1^3)\approx-4.0000\ \checkmark;\quad \text{line passes through the point at } t_1\ \checkmark\]
\[t_0\approx-2.60267\text{: same two checks}\ \checkmark\]
Result
Final answer:
\[\boxed{\,y = \frac{2-t_{0}^{3}}{2}\,x + \frac{t_{0}^{3}+3t_{0}-2}{2}, \qquad t_{0}^{3} = -7\pm\sqrt{113}\,}\]
Final answer:
\[\boxed{\,t_{0}^{6} + 14\,t_{0}^{3} - 64 = 0\,}\]
and the corresponding line is $y=\dfrac{2-t_0^3}{2}x+\dfrac{t_0^3+3t_0-2}{2}$ for each of the two real roots $t_0$.
Remark
Remark. Unlike Problem 529, this curve's version of the “tangent-equals-normal” condition does not resolve into a nice radical: the governing equation is only a quadratic in $u=t_0^3$, but its roots are irrational and the cube roots $t_0=\sqrt[3]{-7\pm\sqrt{113}}$ do not simplify further. The mechanism (a double-root condition combined with a perpendicularity condition) is exactly the one used in Problem 529; it is the curve's own algebra, not the method, that determines whether the final numbers are tidy.