If the point (3, 4) lies on the locus of the point of intersection of the lines `x cos alpha + y sin alpha`=a and `x sin alpha - y cos alpha =b` (`alpha` is a variable), the point (a, b) lies on the line `3x-4y=0` then `(a^(2)-b^(2))/2` is equal to

Answer

To solve the problem step by step, we will analyze the given equations and conditions.

### Step 1: Understand the intersection of the lines
We have two lines given by:
1. \( x \cos \alpha + y \sin \alpha = a \)
2. \( x \sin \alpha - y \cos \alpha = b \)

Let's denote the point of intersection of these lines as \( (h, k) \). We need to find the locus of this point as \( \alpha \) varies.

### Step 2: Solve for \( h \) and \( k \)
To find \( h \) and \( k \), we can manipulate the equations.

From the first equation:
\[ h \cos \alpha + k \sin \alpha = a \quad \text{(1)} \]

From the second equation:
\[ h \sin \alpha - k \cos \alpha = b \quad \text{(2)} \]

### Step 3: Multiply and rearrange
Multiply equation (1) by \( \sin \alpha \) and equation (2) by \( \cos \alpha \):
\[ h \cos \alpha \sin \alpha + k \sin^2 \alpha = a \sin \alpha \quad \text{(3)} \]
\[ h \sin^2 \alpha - k \cos^2 \alpha = b \cos \alpha \quad \text{(4)} \]

### Step 4: Add equations (3) and (4)
Adding equations (3) and (4):
\[ h \cos \alpha \sin \alpha + k \sin^2 \alpha + h \sin^2 \alpha - k \cos^2 \alpha = a \sin \alpha + b \cos \alpha \]
This simplifies to:
\[ h (\cos \alpha + \sin \alpha) + k (\sin^2 \alpha - \cos^2 \alpha) = a \sin \alpha + b \cos \alpha \]

### Step 5: Find the locus
To find the locus, we can express \( h \) and \( k \) in terms of \( a \) and \( b \). We can use the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \) to simplify our expressions.

### Step 6: Substitute the point (3, 4)
We know that the point \( (3, 4) \) lies on the locus, so we substitute \( h = 3 \) and \( k = 4 \):
\[ 3^2 + 4^2 = a^2 + b^2 \]
This gives:
\[ 9 + 16 = a^2 + b^2 \implies a^2 + b^2 = 25 \quad \text{(5)} \]

### Step 7: Use the line condition
The point \( (a, b) \) lies on the line \( 3x - 4y = 0 \), which implies:
\[ 3a - 4b = 0 \implies 3a = 4b \implies a = \frac{4}{3}b \quad \text{(6)} \]

### Step 8: Substitute equation (6) into equation (5)
Now substitute \( a = \frac{4}{3}b \) into \( a^2 + b^2 = 25 \):
\[ \left(\frac{4}{3}b\right)^2 + b^2 = 25 \]
\[ \frac{16}{9}b^2 + b^2 = 25 \]
\[ \frac{16}{9}b^2 + \frac{9}{9}b^2 = 25 \]
\[ \frac{25}{9}b^2 = 25 \]
\[ b^2 = 9 \implies b = 3 \quad \text{(since } b \text{ is positive)} \]

### Step 9: Find \( a \)
Using \( b = 3 \) in equation (6):
\[ a = \frac{4}{3} \times 3 = 4 \]

### Step 10: Calculate \( \frac{a^2 - b^2}{2} \)
Now we need to calculate:
\[ \frac{a^2 - b^2}{2} = \frac{4^2 - 3^2}{2} = \frac{16 - 9}{2} = \frac{7}{2} \]

### Final Answer
Thus, the value of \( \frac{a^2 - b^2}{2} \) is:
\[ \frac{7}{2} \]