The sum of the length of tangent and subtangent at a point of $\mathrm{y}=\mathrm{alog}({\mathrm{x}}^2-{\mathrm{a}}^2)(\mathrm{a}>0)$ is proportional to

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{{\mathrm{x}}^2-{\mathrm{a}}^2}\left(2\mathrm{x}\right)=\frac{2\mathrm{ax}}{{\mathrm{x}}^2-{\mathrm{a}}^2}=\mathrm{m} \\ \mathrm{L}.\mathrm{T}=\frac{\mathrm{y}\sqrt{1+{\mathrm{m}}^2}}{\mathrm{m}}=\frac{\mathrm{y}\sqrt{1+{\displaystyle \frac{4{\mathrm{a}}^2{\mathrm{x}}^2}{{({\mathrm{x}}^2-{\mathrm{a}}^2)}^2}}}}{{\displaystyle \frac{2\mathrm{ax}}{{\mathrm{x}}^2-{\mathrm{a}}^2}}} \\ =\frac{\mathrm{y}({\mathrm{x}}^2+{\mathrm{a}}^2)}{2\mathrm{ax}} \\ \mathrm{L}.\mathrm{S}.\mathrm{T}=\left|\frac{\mathrm{y}}{\mathrm{m}}\right|=\frac{\mathrm{y}({\mathrm{x}}^2-{\mathrm{a}}^2)}{2\mathrm{ax}} \\ \mathrm{L}.\mathrm{T}+\mathrm{L}.\mathrm{S}.\mathrm{T}=\frac{\mathrm{y}}{2\mathrm{ax}}\left({\mathrm{x}}^2+{\mathrm{a}}^2+{\mathrm{x}}^2-{\mathrm{a}}^2\right)=\frac{\mathrm{xy}}{\mathrm{a}} \\ \propto \mathrm{xy} \end{gathered}$$