Show that for the curve y=a ln(x²-a²), the sum

Show that for the curve $y=a\ln(x^2-a^2)$, the sum of the tangent length and the subtangent is proportional to the product $xy$.

Step 1: The two lengths

For $y=f(x)$, the subtangent is $T=y_0/y_0'$ and the length of the tangent segment is $T_L=T\sqrt{1+(y_0')^2}=\dfrac{y_0}{y_0'}\sqrt{1+(y_0')^2}$.

Step 2: Differentiate

\[ y' = \frac{2ax}{x^2-a^2}. \tag{1} \]

At $(x_0,y_0)$, taking $x_0>a>0$ so that $x_0^2-a^2>0$: \[y_0' = \frac{2ax_0}{x_0^2-a^2}, \qquad \frac{1}{y_0'} = \frac{x_0^2-a^2}{2ax_0}. \tag{2}\]

Step 3: Simplify $\sqrt{1+(y_0')^2}$

\[ 1+(y_0')^2 = 1+\frac{4a^2x_0^2}{(x_0^2-a^2)^2} = \frac{(x_0^2-a^2)^2+4a^2x_0^2}{(x_0^2-a^2)^2} = \frac{(x_0^2+a^2)^2}{(x_0^2-a^2)^2}, \tag{3} \] using $(x_0^2-a^2)^2+4a^2x_0^2=x_0^4-2a^2x_0^2+a^4+4a^2x_0^2=(x_0^2+a^2)^2$. Since $x_0>a>0$, both factors are positive, so

\[ \sqrt{1+(y_0')^2} = \frac{x_0^2+a^2}{x_0^2-a^2}. \tag{4} \]

Step 4: Subtangent and tangent length

\[ T = \frac{y_0}{y_0'} = y_0\cdot\frac{x_0^2-a^2}{2ax_0}, \qquad T_L = T\sqrt{1+(y_0')^2} = y_0\cdot\frac{x_0^2-a^2}{2ax_0}\cdot\frac{x_0^2+a^2}{x_0^2-a^2} = y_0\cdot\frac{x_0^2+a^2}{2ax_0}. \tag{5} \]

Step 5: The sum

\[ T_L+T = \frac{y_0(x_0^2+a^2)+y_0(x_0^2-a^2)}{2ax_0} = \frac{2x_0^2y_0}{2ax_0} = \frac{x_0y_0}{a}. \tag{6} \]

Step 6: Verification

\[a=1.5,\ x_0=4\text{: } y_0=a\ln(x_0^2-a^2)\approx3.51713,\ T_L+T \text{ computed directly } = 9.37901,\ x_0y_0/a=9.37901\ \checkmark\]