Determine the average length of all vertical chords of the hyperbola `(x^(2))/(a^(2)) - (y^(2))/(b^(2))=1` over the interval `a le x le 2a`

Answer

To determine the average length of all vertical chords of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) over the interval \(a \leq x \leq 2a\), we can follow these steps:

### Step 1: Find the equation of the hyperbola
The hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
From this equation, we can express \(y\) in terms of \(x\):
\[ \frac{y^2}{b^2} = \frac{x^2}{a^2} - 1 \implies y^2 = b^2\left(\frac{x^2}{a^2} - 1\right) \implies y = \pm b\sqrt{\frac{x^2}{a^2} - 1} \]

### Step 2: Determine the length of a vertical chord
For a vertical chord at a specific \(x\), the endpoints are given by \(y = b\sqrt{\frac{x^2}{a^2} - 1}\) and \(y = -b\sqrt{\frac{x^2}{a^2} - 1}\). The length \(L\) of the vertical chord at this \(x\) is:
\[ L = 2b\sqrt{\frac{x^2}{a^2} - 1} \]

### Step 3: Set up the integral for average length
To find the average length of the vertical chords over the interval \([a, 2a]\), we need to integrate the length function \(L\) over this interval and then divide by the length of the interval:
\[ \text{Average Length} = \frac{1}{2a - a} \int_a^{2a} L \, dx = \frac{1}{a} \int_a^{2a} 2b\sqrt{\frac{x^2}{a^2} - 1} \, dx \]

### Step 4: Simplify the integral
Substituting \(L\) into the integral:
\[ \text{Average Length} = \frac{2b}{a} \int_a^{2a} \sqrt{\frac{x^2}{a^2} - 1} \, dx \]
Let \(u = \frac{x}{a}\), then \(dx = a \, du\) and the limits change from \(x = a\) to \(u = 1\) and from \(x = 2a\) to \(u = 2\):
\[ = \frac{2b}{a} \int_1^2 \sqrt{u^2 - 1} \cdot a \, du = 2b \int_1^2 \sqrt{u^2 - 1} \, du \]

### Step 5: Evaluate the integral
The integral \(\int \sqrt{u^2 - 1} \, du\) can be evaluated using the formula:
\[ \int \sqrt{u^2 - 1} \, du = \frac{u}{2}\sqrt{u^2 - 1} - \frac{1}{2}\ln\left(u + \sqrt{u^2 - 1}\right) + C \]
Evaluating from \(1\) to \(2\):
\[ = \left[ \frac{u}{2}\sqrt{u^2 - 1} - \frac{1}{2}\ln\left(u + \sqrt{u^2 - 1}\right) \right]_1^2 \]
Calculating at the limits:
1. At \(u = 2\):
\[ = \frac{2}{2}\sqrt{2^2 - 1} - \frac{1}{2}\ln\left(2 + \sqrt{3}\right) = \sqrt{3} - \frac{1}{2}\ln(2 + \sqrt{3}) \]
2. At \(u = 1\):
\[ = \frac{1}{2}\sqrt{1^2 - 1} - \frac{1}{2}\ln(1 + 0) = 0 \]

Thus, the integral evaluates to:
\[ = \sqrt{3} - 0 - \left(-\frac{1}{2}\ln(1)\right) = \sqrt{3} - 0 \]

### Step 6: Final average length calculation
Putting it all together:
\[ \text{Average Length} = 2b \left(\sqrt{3} - \frac{1}{2}\ln(2 + \sqrt{3})\right) \]

### Final Answer
The average length of all vertical chords of the hyperbola over the interval \(a \leq x \leq 2a\) is:
\[ \text{Average Length} = 2b \left(\sqrt{3} - \frac{1}{2}\ln(2 + \sqrt{3})\right) \]