The lines parallel to the normal to the curve `x y=1` is/are (a)`3x+4y+5=0` (b) `3x-4y+5=0` (c)`4x+3y+5=0` (d) `3y-4x+5=0`

Options

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

Explanation

To find the lines parallel to the normal to the curve given by the equation \( xy = 1 \), we will follow these steps:

### Step 1: Differentiate the curve
We start with the equation of the curve:
\[ xy = 1 \]
Differentiating both sides with respect to \( x \) using the product rule gives:
\[ y + x \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation from Step 1, we find:
\[ \frac{dy}{dx} = -\frac{y}{x} \]
This represents the slope of the tangent line to the curve at any point.

### Step 3: Find the slope of the normal
The slope of the normal line (\( m_l \)) is the negative reciprocal of the slope of the tangent line (\( m_t \)):
\[ m_l = -\frac{1}{m_t} = -\frac{1}{-\frac{y}{x}} = \frac{x}{y} \]

### Step 4: Substitute \( y \) in terms of \( x \)
From the curve equation \( xy = 1 \), we can express \( y \) as:
\[ y = \frac{1}{x} \]
Substituting this into the slope of the normal gives:
\[ m_l = \frac{x}{\frac{1}{x}} = x^2 \]

### Step 5: Analyze the slope of the normal
Since \( x^2 \) is always greater than 0 for \( x \neq 0 \), the slope of the normal line is always positive.

### Step 6: Check the options for parallel lines
Now, we check the given options to find lines with a positive slope:

1. **Option (a)**: \( 3x + 4y + 5 = 0 \)
- Rearranging gives \( y = -\frac{3}{4}x - \frac{5}{4} \) (slope = -3/4, negative)

2. **Option (b)**: \( 3x - 4y + 5 = 0 \)
- Rearranging gives \( y = \frac{3}{4}x + \frac{5}{4} \) (slope = 3/4, positive)

3. **Option (c)**: \( 4x + 3y + 5 = 0 \)
- Rearranging gives \( y = -\frac{4}{3}x - \frac{5}{3} \) (slope = -4/3, negative)

4. **Option (d)**: \( 3y - 4x + 5 = 0 \)
- Rearranging gives \( y = \frac{4}{3}x - \frac{5}{3} \) (slope = 4/3, positive)

### Conclusion
The lines that are parallel to the normal to the curve \( xy = 1 \) are:
- Option (b): \( 3x - 4y + 5 = 0 \)
- Option (d): \( 3y - 4x + 5 = 0 \)

### Final Answer
The correct options are (b) and (d).

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