To solve the problem step by step, let's break it down into manageable parts:
### Step 1: Understanding the Given Lines
We have two lines:
1. \(3x - 4y = \frac{4}{\pi} \sin^{-1}(a^8 + 1) + 2 \cos^{-1}(a^{12} + 1) - \sec^{-1}(a^2 + 1)\)
2. \(6x - 8y = 7\)
### Step 2: Finding the Value of \(a\)
We need to analyze the expressions involving inverse trigonometric functions to find the value of \(a\).
- For \(\sin^{-1}(x)\), the domain is \([-1, 1]\).
- For \(\cos^{-1}(x)\), the domain is also \([-1, 1]\).
- For \(\sec^{-1}(x)\), the domain is \((-\infty, -1] \cup [1, \infty)\).
Now, we analyze the first term:
\[
\sin^{-1}(a^8 + 1)
\]
For this to be valid, we need:
\[
-1 \leq a^8 + 1 \leq 1
\]
This simplifies to:
\[
-2 \leq a^8 \leq 0
\]
Since \(a^8\) is always non-negative, the only solution is \(a = 0\).
Next, we check the second term:
\[
\cos^{-1}(a^{12} + 1)
\]
For this term:
\[
0 \leq a^{12} + 1 \leq 1
\]
This simplifies to:
\[
0 \leq a^{12} \leq 0
\]
Again, the only solution is \(a = 0\).
### Step 3: Substituting \(a = 0\) into the Lines
Now substitute \(a = 0\) into the equations of the lines:
1. For the first line:
\[
3x - 4y = \frac{4}{\pi} \sin^{-1}(1) + 2 \cos^{-1}(1) - \sec^{-1}(1)
\]
Calculating these values:
- \(\sin^{-1}(1) = \frac{\pi}{2}\)
- \(\cos^{-1}(1) = 0\)
- \(\sec^{-1}(1) = 0\)
Thus, we have:
\[
3x - 4y = \frac{4}{\pi} \cdot \frac{\pi}{2} + 2 \cdot 0 - 0 = 2
\]
So, the first line becomes:
\[
3x - 4y = 2
\]
2. The second line remains:
\[
6x - 8y = 7
\]
### Step 4: Checking if the Lines are Parallel
To check if the lines are parallel, we can rewrite them in slope-intercept form:
1. From \(3x - 4y = 2\):
\[
4y = 3x - 2 \implies y = \frac{3}{4}x - \frac{1}{2}
\]
Slope = \(\frac{3}{4}\)
2. From \(6x - 8y = 7\):
\[
8y = 6x - 7 \implies y = \frac{3}{4}x - \frac{7}{8}
\]
Slope = \(\frac{3}{4}\)
Since both lines have the same slope, they are indeed parallel.
### Step 5: Finding the Distance Between the Lines
The distance \(d\) between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) can be calculated using:
\[
d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}
\]
For our lines:
- First line: \(3x - 4y - 2 = 0\) (so \(C_1 = -2\))
- Second line: \(6x - 8y - 7 = 0\) (so \(C_2 = -7\))
Rewriting the second line in the same form:
\[
3x - 4y - \frac{7}{2} = 0 \implies C_2 = -\frac{7}{2}
\]
Calculating the distance:
\[
d = \frac{\left| -\frac{7}{2} + 2 \right|}{\sqrt{3^2 + (-4)^2}} = \frac{\left| -\frac{7}{2} + \frac{4}{2} \right|}{\sqrt{9 + 16}} = \frac{\left| -\frac{3}{2} \right|}{5} = \frac{3/2}{5} = \frac{3}{10}
\]
### Step 6: Finding the Radius of the Circle
The distance \(d\) we calculated is the diameter of the circle, so the radius \(r\) is:
\[
r = \frac{d}{2} = \frac{3}{20}
\]
### Step 7: Finding the Length of the Arc
The length of the arc \(L\) of a circle is given by:
\[
L = r \theta
\]
where \(\theta\) is in radians. Here, \(\theta = \frac{40}{3}\).
Calculating the length of the arc:
\[
L = \frac{3}{20} \cdot \frac{40}{3} = 2
\]
### Final Answer
The length of the arc is \(2\).
