Problem source

In this question, $r$ and $\theta$ are polar coordinates with $r \ge0$ and  $- \pi < \theta\le  \pi$,  and $a$ and $b$ are positive constants. 
Let $L$ be a fixed line and let $A$ be a fixed point not lying on $L$. Then  the locus of points that are a fixed distance (call it $d$) from $L$ measured along lines through $A$ is called a \textit{ conchoid of Nicomedes}.
\begin{questionparts}
\item Show that if
\[
\vert r- a \sec\theta \vert = b\,,
\tag{$*$}
\]
where $a>b$, then $\sec\theta >0$.   Show that all points with coordinates satisfying ($*$) lie on a certain conchoid of Nicomedes (you should identify  $L$, $d$ and $A$). Sketch the locus of these points.
\item In the case $a < b$, sketch the curve (including the loop for which $\sec\theta<0$) given by 
\[
\vert r- a \sec\theta \vert = b\,
.
\]
Find the area of the loop in the case $a=1$ and $b=2$.  
                             
[Note: 
$
%\displaystyle 
\int \! \sec\theta \,\d \theta = \ln \vert \sec\theta + \tan\theta \vert + C
\,.
$]
\end{questionparts}

Solution source

\begin{questionparts}
\item $r = a \sec \theta \pm b$. The points on $r = a \sec \theta \Leftrightarrow r \cos \theta = a \Leftrightarrow x = a$ are points on the line $x = a$. Therefore points on the curve $r = a \sec \theta \pm b$ are points which are a distance $b$ from the line $x = a$ measured towards $O$. So $A$ is the origin and $d = b$.


\begin{center}
    \begin{tikzpicture}[scale=1]
        \draw[thick,->,>=latex] (-2,0)--(4,0) node[above] {$x$};
        \draw[thick,->,>=latex] (0,-4)--(0,4) node[left] {$y$};
        \begin{scope}
            % You can modify these values to change your plotting region
            \clip (-2,-4) rectangle (4,4);
            
            \draw[domain=0:89,samples=250, color=red, thick] plot (\x:{1 + 1.5/cos(\x)}); 
            \draw[domain=91:269,samples=250, color=red, thick] plot (\x:{1 + 1.5/cos(\x)}); 
            \draw[domain=271:359,samples=250, color=red, thick] plot (\x:{1 + 1.5/cos(\x)});


            \draw[domain=0:89,samples=250, color=red, thick] plot (\x:{-1 + 1.5/cos(\x)}); 
            \draw[domain=91:269,samples=250, color=red, thick] plot (\x:{-1 + 1.5/cos(\x)}); 
            \draw[domain=271:359,samples=250, color=red, thick] plot (\x:{-1 + 1.5/cos(\x)});
            \filldraw (0,0) circle (1pt) node[above left] {$A$};
        \end{scope}
        \draw[dashed] (1.5, -4) -- (1.5, 4) node[above] {$L$};
    
    \end{tikzpicture}
\end{center}

\item \begin{center}
    \begin{tikzpicture}[scale=1]
        \draw[thick,->,>=latex] (-2,0)--(4,0) node[above] {$x$};
        \draw[thick,->,>=latex] (0,-4)--(0,4) node[left] {$y$};
        \begin{scope}
            % You can modify these values to change your plotting region
            \clip (-2,-4) rectangle (4,4);
            \def\a{1};
            \def\b{2};
            \draw[domain=0:89,samples=250, color=red, thick] plot (\x:{\b + \a/cos(\x)}); 
            \draw[domain=91:269,samples=250, color=red, thick] plot (\x:{\b + \a/cos(\x)}); 
            \draw[domain=271:359,samples=250, color=red, thick] plot (\x:{\b + \a/cos(\x)});


            \draw[domain=0:89,samples=250, color=red, thick] plot (\x:{-\b + \a/cos(\x)}); 
            \draw[domain=91:269,samples=250, color=red, thick] plot (\x:{-\b + \a/cos(\x)}); 
            \draw[domain=271:359,samples=250, color=red, thick] plot (\x:{-\b + \a/cos(\x)});
            \filldraw (0,0) circle (1pt) node[above left] {$A$};
        \end{scope}
        \draw[dashed] (1.5, -4) -- (1.5, 4) node[above] {$L$};
    
    \end{tikzpicture}
\end{center}

The loop starts and ends when $r = a \sec \theta - b = 0 \Rightarrow \cos \theta = \frac{a}{b}$, so when $a = 1, b = 2$, this is $-\frac{\pi}{3}$ to $\frac{\pi}{3}$

\begin{align*}
&& A &= \frac12 \int r^2 \d \theta \\
&&&= \frac12 \int_{-\pi/3}^{\pi/3} \left ( \sec \theta - 2 \right)^2 \d \theta \\
&&&= \frac12 \int_{-\pi/3}^{\pi/3} \left (\sec^2 \theta - 4 \sec \theta + 4\right)\d \theta \\
&&&= \frac12 \left [ \tan \theta -4 \ln | \sec \theta + \tan \theta| + 4 \theta \right]_{-\pi/3}^{\pi/3} \\
&&&= \frac12 \left (\left (\tan \frac{\pi}3 - 4 \ln | \sec  \frac{\pi}3 + \tan \frac{\pi}3 | + 4\left ( \frac{\pi}3 \right)\right) - \left (\tan \left (-\frac{\pi}3 \right) - 4 \ln | \sec  \left (-\frac{\pi}3 \right)+ \tan\left ( -\frac{\pi}3 \right) | + 4\left ( -\frac{\pi}3 \right)\right) \right) \\
&&&= \frac12 \left ( 2\sqrt{3} - 4 \ln |2 + \sqrt{3}| + 4 \ln |2-\sqrt{3}| + \frac{8\pi}3 \right) \\
&&&= \sqrt{3} + 2\ln \frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{4\pi}3 \\
&&&= \sqrt{3} + 4 \ln (2 - \sqrt{3})+ \frac{4\pi}3 
\end{align*}
\end{questionparts}