From the point (3/5, 0), draw tangents to the hyperbola

From the point $\left(\frac{3}{5}, 0\right)$, draw tangents to the hyperbola $x^2 - y^2 = 1$ and find the area bounded by these tangents and the arc of the hyperbola between the tangent points.

We will find the tangent lines, then subtract the area "cut off" by the hyperbola's arc from the triangle the tangents form.

Step 1: Find the tangent points

For a point $(x_0,y_0)$ on the hyperbola $x^2-y^2=1$, the tangent line there is $x_0x - y_0y = 1$. For this line to pass through $\left(\frac35,0\right)$:

\[x_0 \cdot \frac{3}{5} = 1 \implies x_0 = \frac{5}{3}\]

Since $(x_0,y_0)$ is on the hyperbola:

\[\left(\frac{5}{3}\right)^2 - y_0^2 = 1 \implies y_0^2 = \frac{16}{9} \implies y_0 = \pm\frac{4}{3}\]

So the tangent points are $\left(\frac{5}{3},\frac{4}{3}\right)$ and $\left(\frac{5}{3},-\frac{4}{3}\right)$.

Step 2: Find the tangent line equations

Tangent at $\left(\frac53,\frac43\right)$: $\frac53x-\frac43y=1 \implies 5x-4y=3 \implies y=\frac{5x-3}{4}$

Tangent at $\left(\frac53,-\frac43\right)$: $\frac53x+\frac43y=1 \implies 5x+4y=3 \implies y=\frac{3-5x}{4}$

Both lines indeed pass through $\left(\frac35,0\right)$: $5\cdot\frac35=3$. ✓

Step 3: Set up the area

The region bounded by the two tangent lines and the hyperbola's arc is symmetric about the $x$-axis. It consists of the triangle formed by the external point $\left(\frac35,0\right)$ and the two tangent points, with the "cap" between the hyperbola and the chord of tangent points removed from each side:

\[A = A_{\triangle} - 2\int_1^{5/3} \sqrt{x^2-1}\,dx\]

The integral only runs from $x=1$ (the hyperbola's vertex) since the hyperbola does not exist for $x<1$, so it is only over $\left[1,\frac53\right]$ that the arc departs from the straight tangent lines.

Step 4: Calculate the triangle area

The triangle has vertices $\left(\frac35,0\right)$, $\left(\frac53,\frac43\right)$, $\left(\frac53,-\frac43\right)$. Base: $\frac43-\left(-\frac43\right)=\frac83$. Height: $\frac53-\frac35=\frac{16}{15}$.

\[A_{\triangle} = \frac{1}{2}\cdot\frac83\cdot\frac{16}{15} = \frac{64}{45}\]

Step 5: Calculate the area under the hyperbola from $x=1$ to $x=\frac53$

Using the antiderivative from Problem 23:

\[\int_1^{5/3} \sqrt{x^2-1}\,dx = \left[\frac{x\sqrt{x^2-1}}{2} - \frac12\ln(x+\sqrt{x^2-1})\right]_1^{5/3}\]

At $x=\frac53$: $\frac{\frac53\cdot\frac43}{2} - \frac12\ln\left(\frac53+\frac43\right) = \frac{10}{9} - \frac{\ln3}{2}$. At $x=1$: $0$.

\[\int_1^{5/3} \sqrt{x^2-1}\,dx = \frac{10}{9} - \frac{\ln3}{2}\]

Step 6: Total area

\[A = \frac{64}{45} - 2\left(\frac{10}{9}-\frac{\ln3}{2}\right) = \frac{64}{45} - \frac{20}{9} + \ln3 = -\frac{4}{5} + \ln3\]

\[\boxed{A = \ln 3 - \frac{4}{5} \approx 0.299 \text{ square units}}\]

Summary

\[\boxed{A = \ln 3 - \frac{4}{5} \approx 0.299 \text{ square units}}\]

Key insight: The area between two tangents from an external point and the arc they touch is naturally the triangle they form minus the "bite" the curved arc takes out of it.