$$\frac{1}{2}x_1y_1 = \int_0^{y_1} x dy - \frac{1}{2}a$$

\begin{equation}

a = \begin{cases} 2\int_0^{y_1} \sqrt{1+y^2} dy - y_1\sqrt{1+{y_1}^2}, & x_1 \geq 1 \\ -2\int_0^{y_1} \sqrt{1+y^2} dy + y_1\sqrt{1+{y_1}^2}, & x_1 \leq -1\end{cases} \end{equation}

Hyperbolic Sine

\begin{equation}

\frac{da}{dy_1} = \begin{cases} \frac{2 \left(1+{y_1}^2\right) - \left(1+{y_1}^2\right) - {y_1}^2}{\sqrt{1+{y_1}^2}} , & x_1 \geq 1 \\ \frac{-2 \left(1+{y_1}^2\right) + \left(1+{y_1}^2\right) + {y_1}^2}{\sqrt{1+{y_1}^2}} , & x_1 \leq -1\end{cases}

\end{equation}

\begin{equation}

\frac{da}{dy_1} = \begin{cases} \frac{1}{\sqrt{1+{y_1}^2}} , & x_1 \geq 1 \\ -\frac{1}{\sqrt{1+{y_1}^2}} , & x_1 \leq -1\end{cases}

\end{equation}

\begin{equation}

\left(\frac{dy_1}{da}\right)^2 = 1+{y_1}^2

\end{equation}



\begin{equation}

2\frac{dy_1}{da}\frac{d^2y_1}{{da}^2} = 2y_1\frac{dy_1}{da}

\end{equation}

\begin{equation}

\frac{d^2y_1}{{da}^2} = y_1

\end{equation}



\begin{equation}

y_1 = Ae^{r_Aa} + Be^{r_Ba}

\end{equation}



\begin{equation}

y_1 = Ae^{a} + Be^{-a}

\end{equation}



\begin{align}

A^2e^{2a} - 2AB + B^2e^{-2a} &= 1 + A^2e^{2a} + 2AB + B^2e^{-2a} \\

0 &= A + B

\end{align}

\begin{align}

A &= \pm \frac{1}{2} \\

B &= \mp \frac{1}{2}

\end{align}



\begin{align}

A &= \frac{1}{2} \\

B &= - \frac{1}{2}

\end{align}

\begin{equation}

y_1 = \frac{1}{2}\left(e^{a} - e^{-a}\right)

\end{equation}



\begin{equation}

\sinh{a} = \frac{1}{2}\left(e^{a} - e^{-a}\right)

\end{equation}



Hyperbolic Cosine

\begin{equation}

\cosh^2{a} - 1 = \frac{1}{4}\left(e^{a} - e^{-a}\right)^2

\end{equation}

\begin{equation}

\cosh^2{a} = \frac{1}{4}e^{2a} + \frac{1}{2} + \frac{1}{4}e^{-2a}

\end{equation}

\begin{equation}

\cosh{a} = \pm\frac{1}{2}\left( e^{a} + e^{-a}\right)

\end{equation}



\begin{equation}

\cosh{a} = \frac{1}{2}\left( e^{a} + e^{-a}\right)

\end{equation}



Hyperbolic Tangent

\begin{equation}

\tanh{a} = \frac{e^{a} - e^{-a}}{e^{a} + e^{-a}}

\end{equation}



\begin{equation}

\tanh{a} = \frac{e^{2a} - 1}{e^{2a} + 1}

\end{equation}



Reciprocal Hyperbolic Functions

\begin{align}

\,\text{csch}\,{a} &= \frac{2}{e^{a} - e^{-a}} \\

\,\text{sech}\,{a} &= \frac{2}{e^{a} + e^{-a}} \\

\,\coth{a} &= \frac{e^{a} + e^{-a}}{e^{a} - e^{-a}} \\

&= \frac{e^{2a} + 1}{e^{2a} - 1}

\end{align}



Hyperbolic functions are related to the unit hyperbola, given by $x^2 - y^2 = 1$, analogous to the way trigonometric functions are related to the unit circle. Both trigonometric and hyperbolic functions can be used to parameterize their respective unit conics. However, while the unit circle's central angle, the argument taken by trigonometric functions, is indeed what one might consider to be an "angle" in the usual meaning of the word, the hyperbolic angle taken by hyperbolic functions is perhaps less intuitively defined. The hyperbolic angle, $a$, is double the area enclosed by $x^2 - y^2 =1$, the $x$-axis, and a line segment between the origin and a point, $(x_1, y_1) = (\cosh{a}, \sinh{a})$, on the unit hyperbola. ($\sinh{a}$ is the hyperbolic sine function and $\cosh{a}$ is the hyperbolic cosine function of $a$.) It may be interesting to note that one might analogously view the central angle (in radians) of a unit circle to be double the area enclosed by the unit circle, the $x$-axis, and a line segment between the origin and a point on the unit circle.Often, hyperbolic functions are defined in terms of exponential functions. Here, these definitions will be derived through basic differential equations.To begin these derivations, a relation between the hyperbolic angle and coordinates of an arbitrary point, $(x_1, y_1) = (\cosh{a}, \sinh{a})$, on the unit hyperbola must be found. Consider a right triangle formed by the vertices $(0,0)$, $(x_1, y_1)$, and $(x_1, 0)$, having an area of $x_1y_1/2$. Notice that this triangle's area is also given by the subtraction of $a/2$ from $\int_0^{y_1} x dy$. Therefore, the following relation holds between $x_1$, $y_1$, and $a$:Using the equation of the unit hyperbola, $a$ can be solved for in relation to $y_1$.To derive the formula for the hyperbolic sine function, $\sinh{a} = y_1$, equation $(1)$ must be solved for $y_1$ in relation to $a$. Unfortunately, integrating the integrands of equation $(1)$ proves to be extremely difficult, so differentiation is used here instead for an easier derivation.Differentiating implicitly with respect to $a$ gives:The above equation is a homogeneous second-order linear differential equation and has the general solution:where $A$ and $B$ are constants while $r_A$ and $r_B$ are the two solutions to the equation $r^2=1$.To solve for $A$ and $B$, apply equation $(4)$ and the initial condition stating that $y_1=0$ when $a=0$.By convention, hyperbolic functions parameterize only the right side of the unit hyperbola, for $x\geq1$, meaning $y_1$ and $a$ should share the same sign. Therefore,Since $\sinh{a}=y_1$, the hyperbolic sine function, written in terms of exponential functions, is thus derived to be:Although the hyperbolic cosine function can also be derived in a similar manner through differential equations, a much easier derivation is possible using the already found hyperbolic sine function. Since $\sinh{a}=y_1$ and $\cosh{a} = x_1$, the equation of the unit hyperbola can be used to relate $\cosh{a}$ to the exponential expression that gives $\sinh{a}$.Once again, hyperbolic functions, by convention, parameterize the unit hyperbola only for $x\geq1$, meaning $\cosh{a} = x_1$ should always be greater than or equal to $1$.The hyperbolic tangent function, analogous to the trigonometric tangent function, is equal to $y_1/x_1 = \sinh{a}/\cosh{a}$. Therefore, the hyperbolic tangent function is found to be:or, written in another form,The hyperbolic cosecant, secant, and cotangent functions are simply the reciprocals of the hyperbolic sine, cosine, and tangent functions respectively and are respectively thus found to be: