The equation of a rectangular hyperbola whose asymptotes are `x=3` and `y=5` and passing through (7,8) is (a) `x y-3y+5x+3=0` (b) `x y+3y+4x+3=0` (c) `x y-3y+5x-3=0` (d) `x y-3y+5x+3=0`

Options

A. `xy-3y+5x+3=0`
B. `xy+3y+4x+3=0`
C. `xy-3y+5x-3=0`
D. `xy-3y-5x+3=0`

Explanation

To find the equation of a rectangular hyperbola with given asymptotes and passing through a specific point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the asymptotes**: The asymptotes given are \( x = 3 \) and \( y = 5 \). We can express these as:
\[ x - 3 = 0 \quad \text{(Asymptote 1)} \]
\[ y - 5 = 0 \quad \text{(Asymptote 2)} \]

2. **Form the combined equation of the asymptotes**: The combined equation of the asymptotes can be formed by multiplying the two equations:
\[ (x - 3)(y - 5) = 0 \]
Expanding this gives:
\[ xy - 5x - 3y + 15 = 0 \]

3. **Form the equation of the hyperbola**: The equation of the hyperbola differs from the combined equation of the asymptotes by a constant \( \lambda \):
\[ xy - 5x - 3y + 15 + \lambda = 0 \]

4. **Substitute the point (7, 8)**: Since the hyperbola passes through the point (7, 8), we substitute \( x = 7 \) and \( y = 8 \) into the equation:
\[ 7 \cdot 8 - 5 \cdot 7 - 3 \cdot 8 + 15 + \lambda = 0 \]
Simplifying this:
\[ 56 - 35 - 24 + 15 + \lambda = 0 \]
\[ 56 + 15 = 71 \]
\[ 71 - 35 - 24 + \lambda = 0 \]
\[ 71 - 59 + \lambda = 0 \]
\[ 12 + \lambda = 0 \]
Thus, we find:
\[ \lambda = -12 \]

5. **Substitute \( \lambda \) back into the hyperbola equation**: Now we substitute \( \lambda \) back into the equation of the hyperbola:
\[ xy - 5x - 3y + 15 - 12 = 0 \]
This simplifies to:
\[ xy - 5x - 3y + 3 = 0 \]

6. **Final equation**: The equation of the rectangular hyperbola is:
\[ xy - 5x - 3y + 3 = 0 \]

### Conclusion:
The equation of the rectangular hyperbola whose asymptotes are \( x = 3 \) and \( y = 5 \), and that passes through the point (7, 8) is:
\[ xy - 5x - 3y + 3 = 0 \]