Show that the length of the tangent of the curve

Show that the length of the tangent of the curve $x=\sqrt{a^2-y^2}-a\ln\!\left(\dfrac{a+\sqrt{a^2-y^2}}{y}\right)$ is constant.

Step 1: The tangent-length formula, with $x$ as a function of $y$

For a curve given as $x=x(y)$, the length of the tangent at $(x,y)$ is $T=y\sqrt{1+(dx/dy)^2}$.

Step 2: A substitution

Let $u=\sqrt{a^2-y^2}$, so $\dfrac{du}{dy}=\dfrac{-y}{u}$, and the curve reads $x=u-a\bigl[\ln(a+u)-\ln y\bigr]$.

Step 3: Differentiate

\[ \frac{dx}{dy} = \frac{du}{dy} - a\left[\frac{1}{a+u}\cdot\frac{du}{dy}-\frac{1}{y}\right] = -\frac{y}{u}+\frac{ay}{u(a+u)}+\frac{a}{y}. \tag{1} \]

Combine the first two terms over $u(a+u)$: \[-\frac{y}{u}+\frac{ay}{u(a+u)} = \frac{-y(a+u)+ay}{u(a+u)} = \frac{-yu}{u(a+u)} = -\frac{y}{a+u}. \tag{2}\]

So \[\frac{dx}{dy} = -\frac{y}{a+u}+\frac{a}{y} = \frac{a(a+u)-y^2}{y(a+u)} = \frac{a^2+au-y^2}{y(a+u)}. \tag{3}\]

Since $u^2=a^2-y^2$, the numerator is $u^2+au=u(u+a)$, so

\[ \frac{dx}{dy} = \frac{u(u+a)}{y(a+u)} = \frac{u}{y} = \frac{\sqrt{a^2-y^2}}{y}. \tag{4} \]

Step 4: Compute the tangent length

\[ 1+\left(\frac{dx}{dy}\right)^2 = 1+\frac{a^2-y^2}{y^2} = \frac{a^2}{y^2} \;\Longrightarrow\; T = y\cdot\frac{a}{y} = a. \tag{5} \]

Step 5: Verification

\[a=2,\ y=1\text{: } dx/dy=\sqrt{3}\approx1.73205\ (\text{matching the closed form (4) to machine precision}),\ T=a=2.00000\ \checkmark\]

\[a=5,\ y=0.3\text{: } T \text{ computed from the raw derivative}=5.00000=a\ \checkmark\]