The family of curves whose tangent form an angle `(pi)/(4)` with the hyperbola xy=1,is

Options

A. `y=x-2tan^(-1)(x)+k`
B. `y=x+2tan^(-1)(x)+k`
C. `y=2x-tan^(-1)(x)+k`
D. `y=2x+tan^(-1)(x)+k`

Answer

To solve the problem of finding the family of curves whose tangents form an angle \(\frac{\pi}{4}\) with the hyperbola \(xy = 1\), we can follow these steps:

### Step 1: Understand the given hyperbola
The hyperbola is given by the equation:
\[ xy = 1 \]
We can express \(y\) in terms of \(x\):
\[ y = \frac{1}{x} \]

### Step 2: Find the slope of the hyperbola
To find the slope of the tangent to the hyperbola, we differentiate \(y = \frac{1}{x}\) with respect to \(x\):
\[ \frac{dy}{dx} = -\frac{1}{x^2} \]
Thus, the slope \(m_1\) of the tangent to the hyperbola is:
\[ m_1 = -\frac{1}{x^2} \]

### Step 3: Set up the angle condition
Let the slope of the tangent to the family of curves be \(m_2\). The angle \(\theta\) between the two tangents is given by:
\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]
Since we know \(\theta = \frac{\pi}{4}\), we have:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, we can set up the equation:
\[ 1 = \left| \frac{m_2 + \frac{1}{x^2}}{1 - \frac{1}{x^2} m_2} \right| \]

### Step 4: Solve the equation
This gives us two cases to consider:
1. \(\frac{m_2 + \frac{1}{x^2}}{1 - \frac{1}{x^2} m_2} = 1\)
2. \(\frac{m_2 + \frac{1}{x^2}}{1 - \frac{1}{x^2} m_2} = -1\)

#### Case 1:
From the first case:
\[ m_2 + \frac{1}{x^2} = 1 - \frac{1}{x^2} m_2 \]
Rearranging gives:
\[ m_2 + \frac{1}{x^2} m_2 = 1 - \frac{1}{x^2} \]
\[ m_2(1 + \frac{1}{x^2}) = 1 - \frac{1}{x^2} \]
Thus,
\[ m_2 = \frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} = \frac{x^2 - 1}{x^2 + 1} \]

#### Case 2:
From the second case:
\[ m_2 + \frac{1}{x^2} = -1 + \frac{1}{x^2} m_2 \]
Rearranging gives:
\[ m_2 + \frac{1}{x^2} m_2 = -1 - \frac{1}{x^2} \]
\[ m_2(1 + \frac{1}{x^2}) = -1 - \frac{1}{x^2} \]
Thus,
\[ m_2 = \frac{-1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} = \frac{-x^2 - 1}{x^2 + 1} \]

### Step 5: Integrate to find the family of curves
We will consider the first case for integration:
\[ \frac{dy}{dx} = \frac{x^2 - 1}{x^2 + 1} \]
Rearranging gives:
\[ dy = \frac{x^2 - 1}{x^2 + 1} dx \]

Now, integrate both sides:
\[ y = \int \frac{x^2 - 1}{x^2 + 1} dx \]

### Step 6: Perform the integration
We can split the integral:
\[ y = \int \left(1 - \frac{2}{x^2 + 1}\right) dx \]
This results in:
\[ y = x - 2 \tan^{-1}(x) + C \]

### Final Result
Thus, the family of curves is given by:
\[ y = x - 2 \tan^{-1}(x) + k \]
where \(k\) is the constant of integration.