Find the area bounded by the arc of the hyperbola x² - y²

Find the area bounded by the arc of the hyperbola $x^2 - y^2 = 1$, the $x$-axis, and the ordinate (vertical line) through an arbitrary point.

We will express the area as a function $A(t)$ of the vertical line's position.

Step 1: Understand the problem

The hyperbola $x^2 - y^2 = 1$ has two branches: the right branch ($x\geq1$) and the left branch ($x\leq-1$). We work with the right branch. For a point $(t, y)$ on the hyperbola with $t>1$, we want the area bounded by the hyperbola from $(1,0)$ to $(t,y)$, the $x$-axis from $(1,0)$ to $(t,0)$, and the vertical line (ordinate) at $x=t$.

Step 2: Express $y$ in terms of $x$

From $x^2-y^2=1$: $y^2 = x^2-1$, so for the upper branch ($y\geq0$):

\[y = \sqrt{x^2-1}\]

Step 3: Set up the integral

\[A(t) = \int_1^t \sqrt{x^2 - 1} \, dx\]

Step 4: Evaluate using the standard antiderivative

The standard result (derivable via the hyperbolic substitution $x=\cosh u$) is:

\[\int \sqrt{x^2 - 1} \, dx = \frac{x\sqrt{x^2-1}}{2} - \frac{1}{2}\ln|x + \sqrt{x^2-1}| + C\]

Therefore:

\[A(t) = \left[\frac{x\sqrt{x^2-1}}{2} - \frac{1}{2}\ln(x + \sqrt{x^2-1})\right]_1^t\]

At $x=t$: $\frac{t\sqrt{t^2-1}}{2} - \frac{1}{2}\ln(t+\sqrt{t^2-1})$. At $x=1$: $0 - \frac12\ln(1) = 0$.

\[\boxed{A(t) = \frac{1}{2}\left[t\sqrt{t^2-1} - \ln(t + \sqrt{t^2-1})\right]}\]

Equivalently, in terms of the inverse hyperbolic cosine, $\ln(t+\sqrt{t^2-1}) = \text{arcosh}\,t$, so:

\[\boxed{A(t) = \frac{1}{2}\left[t\sqrt{t^2-1} - \text{arcosh}\, t\right]}\]

Step 5: Derivation check via hyperbolic substitution

Substituting $x=\cosh u$, $dx=\sinh u\,du$, with $u=0$ at $x=1$ and $u=\text{arcosh}\,t$ at $x=t$:

\[A(t) = \int_0^{\text{arcosh}\,t} \sinh^2 u \, du = \frac{1}{2}\left[\frac{\sinh 2u}{2} - u\right]_0^{\text{arcosh}\,t}\]

Using $\sinh(2u)=2\sinh u\cosh u = 2t\sqrt{t^2-1}$ at $u=\text{arcosh}\,t$, this reduces to the same result. ✓

Step 6: Numerical example

For $t=2$:

\[A(2) = \frac{1}{2}\left[2\sqrt{3} - \ln(2+\sqrt{3})\right] \approx \frac12[3.464-1.317]\approx1.074\]

Summary

\[\boxed{A(t) = \frac{1}{2}\left[t\sqrt{t^2-1} - \ln(t + \sqrt{t^2-1})\right]}\]

Key insight: The area formula involves both algebraic and logarithmic terms, reflecting the hyperbolic nature of the curve — analogous to how circular sectors involve inverse trigonometric functions.