The equation of the tangents to the curve `(1+x^(2))y=1` at the points of its intersection with the curve `(x+1)y=1`, is given by

Options

A. `x+y=1, y=1`
B. `x+2y=2, y=1`
C. `x-y=1, y=1`
D. None of these

Answer

To find the equation of the tangents to the curve \( (1 + x^2)y = 1 \) at the points of intersection with the curve \( (x + 1)y = 1 \), we will follow these steps:

### Step 1: Find the points of intersection of the two curves.
We start with the two equations:
1. \( (1 + x^2)y = 1 \)
2. \( (x + 1)y = 1 \)

Rearranging both equations for \( y \):
- From the first equation:
\[ y = \frac{1}{1 + x^2} \]
- From the second equation:
\[ y = \frac{1}{x + 1} \]

Setting these two expressions for \( y \) equal to each other:
\[ \frac{1}{1 + x^2} = \frac{1}{x + 1} \]

### Step 2: Solve the equation for \( x \).
Cross-multiplying gives:
\[ 1 + x^2 = 1 + x \]
This simplifies to:
\[ x^2 = x \]
Factoring out \( x \):
\[ x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

### Step 3: Find the corresponding \( y \) values.
For \( x = 0 \):
\[ y = \frac{1}{1 + 0^2} = 1 \quad \Rightarrow \quad (0, 1) \]
For \( x = 1 \):
\[ y = \frac{1}{1 + 1} = \frac{1}{2} \quad \Rightarrow \quad (1, \frac{1}{2}) \]

The points of intersection are \( (0, 1) \) and \( (1, \frac{1}{2}) \).

### Step 4: Find the slope of the tangent at each point.
We differentiate the first curve \( (1 + x^2)y = 1 \) implicitly:
\[ \frac{d}{dx}[(1 + x^2)y] = \frac{d}{dx}[1] \]
Using the product rule:
\[ (1 + x^2)\frac{dy}{dx} + 2xy = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{2xy}{1 + x^2} \]

### Step 5: Calculate the slope at the point \( (0, 1) \).
Substituting \( x = 0 \) and \( y = 1 \):
\[ \frac{dy}{dx} = -\frac{2(0)(1)}{1 + 0^2} = 0 \]
The slope at \( (0, 1) \) is \( 0 \).

### Step 6: Write the equation of the tangent line at \( (0, 1) \).
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = 0 \), \( (x_1, y_1) = (0, 1) \):
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]

### Step 7: Calculate the slope at the point \( (1, \frac{1}{2}) \).
Substituting \( x = 1 \) and \( y = \frac{1}{2} \):
\[ \frac{dy}{dx} = -\frac{2(1)(\frac{1}{2})}{1 + 1^2} = -\frac{1}{2} \]
The slope at \( (1, \frac{1}{2}) \) is \( -\frac{1}{2} \).

### Step 8: Write the equation of the tangent line at \( (1, \frac{1}{2}) \).
Using the point-slope form:
\[ y - \frac{1}{2} = -\frac{1}{2}(x - 1) \]
Multiplying through by \( 2 \):
\[ 2y - 1 = -x + 1 \quad \Rightarrow \quad x + 2y - 2 = 0 \]

### Final Result:
The equations of the tangents to the curve at the points of intersection are:
1. \( y = 1 \) (at point \( (0, 1) \))
2. \( x + 2y - 2 = 0 \) (at point \( (1, \frac{1}{2}) \))