Animated Solution for Mathematics - Conic Sections: For any real $t$, $x=\frac{e^t+e^{-t}}{2}$, $y=\frac{e^t-e^{-t}}{2}$ is a point on the hyperbola $x^2-y^2=1$. Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to $t_1$ and $-t_1$ is $t_1$.

Welcome, fellow travelers of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a profound connection between algebra and geometry.

We are exploring the rectangular hyperbola, defined by the equation $x^2-y^2=1$. This curve is the hyperbolic counterpart to the unit circle, and just as the circle has its trigonometric functions, the hyperbola has its hyperbolic functions.

We are given the parametric form:

These are not just arbitrary variables; they are the keys to unlocking the area of the hyperbolic sector.

Imagine you are standing at the origin $O(0,0)$ on a Cartesian plane. We have a point $P$ on the hyperbola corresponding to a parameter $t_1$, so $P=(\cosh t_1,\sinh t_1)$.

Now, consider the point $Q$ corresponding to $-t_1$. Because $\cosh (-t_1)=\cosh t_1$ and $\sinh (-t_1)=-\sinh t_1$, the point $Q$ is simply $(\cosh t_1,-\sinh t_1)$.

We are tasked with finding the area bounded by the hyperbola and the lines $OP$ and $OQ$. The region is perfectly symmetric about the $x$-axis, allowing us to calculate the area of the upper sector and multiply it by two.

To find the area of the upper sector, let us drop a perpendicular from $P$ to the $x$-axis at point $R(x_1,0)$. The area we desire is the area of the right-angled triangle $OPR$ minus the area under the hyperbola from the vertex at $x=1$ to $x=x_1$.

The area of the triangle $OPR$ is:

For the area under the curve $A_c$, we integrate the function $y=\sqrt{x^2-1}$:

Using the standard integral formula, we evaluate:

Evaluating at the limits, the lower limit at $x=1$ vanishes, leaving:

Since $P(x_1,y_1)$ lies on the hyperbola, we know that $\sqrt{x_1^2-1}=y_1$. Substituting this into our expression for $A_c$:

Now, return to the area of our upper sector:

The $\frac{1}{2}{x}_1{y}_1$ terms cancel out perfectly, leaving $\frac{1}{2}\ln (x_1+y_1)$. Using our parametric definitions:

Thus, $\ln (x_1+y_1)=\ln (e^{t_1})=t_1$. The area of the upper sector is $\frac{1}{2}{t}_1$.

Multiplying by two for the full sector, we get the total area:

This is a profound result: the parameter $t$ in the hyperbola is not just a coordinate; it is a measure of the area of the hyperbolic sector, just as the angle $\theta$ is a measure of the area of a circular sector. You have just proven a fundamental property of hyperbolic geometry.