We differentiate with respect to \(x\) on both sides, and we have \[ \cosh x + \cosh y \DiffFrac {y}{x} = 0. \]
If the curve has a stationary point \((x, y)\), we must have \(\DiffFrac {y}{x} = 0\) at that point, and hence \[ \cosh x = 0. \]
This is impossible since the \(\cosh \) function has a range of \([1, +\infty )\), and hence \(C\) has no stationary points. (In fact we must have \(\DiffFrac {y}{x} < 0\).)
Differentiating this with respect to \(x\) again gives \[ \sinh x + \sinh y \left (\DiffFrac {y}{x}\right )^2 + \cosh y \NdiffFrac {2}{y}{x} = 0. \]
At point \((x, y)\), \(\NdiffFrac {2}{y}{x} = 0\) if and only if \[ \sinh x + \sinh y \left (\DiffFrac {y}{x}\right )^2 = 0. \]
From the previous differentiation, we know that \[ \DiffFrac {y}{x} = - \frac {\cosh x}{\cosh y}, \] and hence \[ \sinh x + \sinh y \cdot \frac {\cosh ^2 x}{\cosh ^2 y} = 0, \] which gives \[ \cosh ^2 y \sinh x + \sinh y \cosh ^2 x = 0. \]
Using the identity \(\cosh ^2 t = 1 + \sinh ^2 t\), we have \[ \sinh x + \sinh ^2 y \sinh x + \sinh y + \sinh ^2 x \sinh y = 0, \] and hence \[ (\sinh x + \sinh y)(1 + \sinh x \sinh y) = 0. \]
Since \(\sinh x + \sinh y = 2k\) and \(k\) is positive, we can conclude that \[ 1 + \sinh x \sinh y = 0, \] as desired.
The only-if direction is identical since all steps above are reversible.
For a point of inflection, we must first have \(\NdiffFrac {y}{x} = 0\), and hence \[ \sinh x \sinh y = -1, \sinh x + \sinh y = 2k. \]
This means that \(\sinh x\) and \(\sinh y\) are roots to the quadratic equation in \(t\) \[ t^2 - 2kt - 1 = 0. \]
This equation solves to \[ t_{1, 2} = \frac {2k \pm \sqrt {4k^2 + 4}}{2} = k \pm \sqrt {k^2 + 1}. \]
Therefore, the points where the second derivative is zero on the curve are \[ \left (\arsinh \left (k \pm \sqrt {k^2 + 1}\right ), \arsinh \left (k \mp \sqrt {k^2 + 1}\right )\right ). \]
If \(x + y = a\) and \(\sinh x + \sinh y = 2k\), we must have \(y = a - x\), and hence \begin {align*} \frac {e^x - e^{-x}}{2} + \frac {e^{a - x} - e^{x - a}}{2} & = 2k \\ e^{2x} - 1 + e^a - e^{2x - a} & = 4k e^{x} \\ e^{2x} (1 - e^{-a}) - 4k e^{x} + (e^a - 1) & = 0, \end {align*}
as desired.
Since \(e^x\) is always real, we must have \begin {align*} (-4k)^2 - 4 (1 - e^{-a})(e^a - 1) & = 16k^2 - 4 (e^a - 1 - 1 + e^{-a}) \\ & = 16k^2 - 4 (2 \cosh a - 2) \\ & = 16k^2 + 8 - 8 \cosh a \\ & \geq 0, \end {align*}
and hence \[ \cosh a \leq 2k^2 + 1. \]
As for the left-hand side inequality, we already know \(\cosh a \geq 1\). \(\cosh a = 1\) if and only if \(a = x + y = 0\), in which case \[ \sinh x + \sinh y = \sinh x + \sinh (-x) = 0 \neq 2k, \] since \(2k > 0\).
Hence, we must have \[ 1 < \cosh a \leq 2k^2 + 1, \] as desired.
Notice that when \(\cosh a = 2k^2 + 1\), there is precisely one root to the quadratic equation, which means \(x = y\). Hence, \begin {align*} 2k^2 + 1 & = \cosh a \\ & = \cosh (x + y) \\ & = \cosh x \cosh y + \sinh x \sinh y \\ & = \cosh ^2 x + \sinh ^2 x \\ & = 1 + 2 \sinh ^2 x, \end {align*}
which shows that (since \(\sinh x + \sinh y = k\)) \[ \sinh x = \sinh y = k. \]
The graph meets the axis at \((0, \arsinh (2k))\) and \((\arsinh (2k), 0)\).
Hence, the graph must look as follows:
