We prove the first part by contradiction. Assume that \(\sec \theta \ngtr 0\), this means \(\sec \theta \leq -1\).
But in this case, \[ r - a \sec \theta \geq r + a \geq a > b, \] but \(\abs *{r - a \sec \theta } = b\), implies \(r - a \sec \theta \leq b\), and this leads to a contradiction.
This implies that \(\sec \theta > 0\). Hence, \(\cos \theta > 0\), and \(\theta \in \left (-\frac {\pi }{2}, \frac {\pi }{2}\right )\).
We aim to show that \(\abs *{r - a \sec \theta } = b\) lies on the conchoid of Nicomedes where \(L: x = a\) and \(d = b\), with \(A(0, 0)\).
Let \(O\) be the origin, \(P_{\theta }(a, a \tan \theta )\) and \(P_0(a, 0)\). All points on the half-line \(OP_{\theta }\) will have argument \(\theta \).
Let \(Q_{\theta }\) be the points on such line, satisfying the given equation \(\abs *{r - a \sec \theta } = b\).
For every \(\theta \in \left (-\frac {\pi }{2}, \frac {\pi }{2}\right )\), we have \[ \abs *{OP_{\theta }} = \abs *{OP_0} \sec \theta = a \sec \theta . \]
The given equation \(\abs *{r - a \sec \theta } = b\) simplifies to \(r = a \sec \theta \pm b\).
This implies that \(Q_{\theta }\) must lie on the half-line \(OP_{\theta }\) through \(O\), and a fixed distance \(b\) away measured along \(OP_{\theta }\) from line \(L: x = a\) (which is measured from \(P_{\theta }\)).
This is precisely the definition of a conchoid of Nicomedes, and this finishes our proof.
The sketch is as below.
When \(\sec \theta < 0\), \(\sec \theta \leq -1\). We have \(r = a \sec \theta \pm b\).
Since \(r \geq 0\), we must have \(r = a \sec \theta + b \geq 0\) (since if \(r = a \sec \theta - b\), then \(r < 0\)), and hence \[ -1 \geq \sec \theta \geq -\frac {b}{a}, -1 \leq \cos \theta \leq -\frac {a}{b}, \] which means the area of the loop is given by the range of \[ \theta \in \left (-\pi , -\arccos \left (-\frac {a}{b}\right )\right ] \cup \left [\arccos \left (-\frac {a}{b}\right ), \pi \right ]. \]
Therefore, the area of the loop is given by \[ A = \frac {1}{2}\left [\int _{-\pi }^{-\arccos \left (-\frac {a}{b}\right )} r^2 \Diff {\theta } + \int _{\arccos \left (-\frac {a}{b}\right )}^{\pi } r^2 \Diff {\theta }\right ]. \]
Notice that \begin {align*} \int r^2 \Diff \theta & = \int \left (a^2 \sec ^2\theta + 2ab \sec \theta + b^2\right ) \Diff {\theta } \\ & = a^2 \tan \theta + 2ab \ln \abs *{\sec \theta + \tan \theta } + b^2\theta + C \\ & = \tan \theta + 4\ln \abs *{\sec \theta +\tan \theta } + 4\theta + C, \end {align*}
and \[ \alpha = \arccos \left (-\frac {a}{b}\right ) = \arccos \left (-\frac {1}{2}\right ) = \frac {2\pi }{3}. \]
Therefore, \begin {align*} A & = \frac {1}{2}\left [\int _{-\pi }^{-\arccos \left (-\frac {a}{b}\right )} r^2 \Diff {\theta } + \int _{\arccos \left (-\frac {a}{b}\right )}^{\pi } r^2 \Diff {\theta }\right ] \\ & = \frac {1}{2} \left [\left (\tan \theta + 4\ln \abs *{\sec \theta +\tan \theta } + 4\theta \right )^{-\frac {2\pi }{3}}_{-\pi } + \left (\tan \theta + 4\ln \abs *{\sec \theta +\tan \theta } + 4\theta \right )^{\pi }_{\frac {2\pi }{3}}\right ] \\ & = \frac {1}{2} \left [\left (\sqrt {3} + 4\ln \abs *{-2+\sqrt {3}} - \frac {8\pi }{3}\right ) - \left (0 + 4\ln \abs *{-1} - 4\pi \right )\right . \\ & \phantom {=} \left .+ \left (0 + 4\ln \abs *{-1} + 4\pi \right ) - \left (-\sqrt {3} + 4\ln \abs *{-2-\sqrt {3}} + \frac {8\pi }{3}\right )\right ] \\ & = \frac {1}{2} \left (2\sqrt {3} - \frac {16\pi }{3} + 8\pi \right ) + 2\ln (2 - \sqrt {3}) - 2\ln (2 + \sqrt {3}) \\ & = \frac {4}{3}\pi + \sqrt {3} + 2\ln \left (\frac {2 - \sqrt {3}}{2 + \sqrt {3}}\right ) \\ & = \frac {4}{3}\pi + \sqrt {3} + 4\ln (2 - \sqrt {3}). \end {align*}