The tangent at a point P of the curve y = (x-2)²(x+2)³

The tangent at a point $P$ of the curve $y = \bigl(x-2\bigr)^{2}\bigl(x+2\bigr)^{3}$ meets $Ox$ at $T$ and $Oy$ at $T_{1}$; the normal at $P$ meets $Ox$ at $N$ and $Oy$ at $N_{1}$. Prove that $\overline{T_{1}N}:\overline{TN_{1}} = y'$.

Step 1: The four intercepts

Let $P = \bigl(x_{0},y_{0}\bigr)$ and $k = y'\bigl(x_{0}\bigr)\neq0$. From the tangent $y-y_{0} = k\bigl(x-x_{0}\bigr)$ and the normal $y-y_{0} = -\dfrac1k\bigl(x-x_{0}\bigr)$:

\[ T = \left(x_{0}-\frac{y_{0}}{k},\,0\right), \qquad T_{1} = \bigl(0,\,y_{0}-kx_{0}\bigr), \tag{1} \]

\[ N = \bigl(x_{0}+ky_{0},\,0\bigr), \qquad N_{1} = \left(0,\,y_{0}+\frac{x_{0}}{k}\right). \tag{2} \]

Step 2: Two right triangles at the origin

$T_{1}$ and $N_{1}$ lie on $Oy$, while $T$ and $N$ lie on $Ox$; the axes are perpendicular, so by Pythagoras

\[ \overline{T_{1}N}^{\,2} = \overline{ON}^{\,2}+\overline{OT_{1}}^{\,2}, \qquad \overline{TN_{1}}^{\,2} = \overline{OT}^{\,2}+\overline{ON_{1}}^{\,2}. \tag{3} \]

Step 3: The legs differ by the factor $|k|$

Comparing (1) and (2) term by term: \[\overline{ON_{1}} = \left|y_{0}+\frac{x_{0}}{k}\right| = \frac{\bigl|ky_{0}+x_{0}\bigr|}{|k|} = \frac{\overline{ON}}{|k|}, \qquad \overline{OT} = \left|x_{0}-\frac{y_{0}}{k}\right| = \frac{\bigl|kx_{0}-y_{0}\bigr|}{|k|} = \frac{\overline{OT_{1}}}{|k|}. \tag{4}\]

So the second triangle is similar to the first, scaled by $\dfrac{1}{|k|}$.

Step 4: The ratio

Substituting (4) into (3), \[\overline{TN_{1}}^{\,2} = \frac{\overline{OT_{1}}^{\,2}+\overline{ON}^{\,2}}{k^{2}} = \frac{\overline{T_{1}N}^{\,2}}{k^{2}} \;\Longrightarrow\; \frac{\overline{T_{1}N}}{\overline{TN_{1}}} = |k| = \bigl|y'\bigr|. \tag{5}\]

Equivalently, expanding directly: both squares reduce to a common factor, \[\overline{T_{1}N}^{\,2} = \bigl(1+k^{2}\bigr)\bigl(x_{0}^{2}+y_{0}^{2}\bigr), \qquad \overline{TN_{1}}^{\,2} = \frac{1+k^{2}}{k^{2}}\bigl(x_{0}^{2}+y_{0}^{2}\bigr). \tag{6}\]

Step 5: A word on the sign

The two segments join points on different axes, so no orientation makes them signed lengths; what the computation yields is the ratio of ordinary distances, namely $\bigl|y'\bigr|$. The statement $\overline{T_{1}N}:\overline{TN_{1}} = y'$ is therefore to be read with the derivative in absolute value — it holds verbatim only where $y'\gt 0$. At $x_{0} = 1$ on the given curve, for instance, $y' = -27$ while the ratio of lengths is $+27$.

Step 6: The given curve

Nothing above used the particular curve: the identity holds for any differentiable curve at any point with $y'\neq0$ and $P\neq O$. For the curve of the problem, \[y' = 2\bigl(x-2\bigr)\bigl(x+2\bigr)^{3}+3\bigl(x-2\bigr)^{2}\bigl(x+2\bigr)^{2} = \bigl(x-2\bigr)\bigl(x+2\bigr)^{2}\bigl(5x-2\bigr),\] which vanishes only at $x = 2$, $x = -2$ and $x = \tfrac25$; at those three points the normal is vertical, $N_{1}$ does not exist, and the statement is vacuous.

Step 7: Verification

\[x_{0} = 0:\ y_{0} = 32,\ k = 16;\quad T = \bigl(-2,0\bigr),\ T_{1} = \bigl(0,32\bigr),\ N = \bigl(512,0\bigr),\ N_{1} = \bigl(0,32\bigr)\]

\[\overline{T_{1}N} = 512.99903,\quad \overline{TN_{1}} = 32.062439,\quad \text{ratio } = 16.000000 = |k| \;\;\checkmark\]

\[x_{0} = 1:\ k = -27,\ \overline{T_{1}N} = 730,\ \overline{TN_{1}} = 27.037037,\ \text{ratio } = 27 = |k| \;\;\checkmark\]