A family of curves is such that the slope of normal at any point (x, y) is 2(1-y).
The orthogonal trajectories of the given family of curves is

Options

A. (a)`y = ke^(-2x), -1`
B. (b)`y = ke^(2x) + 1`
C. (c)`y = ke^(+2x) - 1`
D. (d)`y = ke^(-2x) + 1`

Answer

To solve the problem, we need to find the orthogonal trajectories of the given family of curves defined by the slope of the normal at any point (x, y) being \(2(1 - y)\).

### Step 1: Understanding the slope of the normal
The slope of the normal at a point (x, y) is given as:
\[ \text{slope of normal} = 2(1 - y) \]
The slope of the tangent line is the negative reciprocal of the slope of the normal. Therefore, we can express the slope of the tangent \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = -\frac{1}{2(1 - y)} \]

### Step 2: Rearranging the equation
We can rearrange the equation to separate variables:
\[ (1 - y) dy = -\frac{1}{2} dx \]

### Step 3: Integrating both sides
Now, we integrate both sides:
\[ \int (1 - y) dy = -\frac{1}{2} \int dx \]
This gives us:
\[ y - \frac{y^2}{2} = -\frac{x}{2} + C \]
where \(C\) is the constant of integration.

### Step 4: Rearranging the equation
Rearranging the equation, we can express it as:
\[ y - \frac{y^2}{2} + \frac{x}{2} = C \]

### Step 5: Finding the orthogonal trajectories
To find the orthogonal trajectories, we need to find the negative reciprocal of the slope of the original family of curves. The slope of the original family of curves is:
\[ \frac{dy}{dx} = -\frac{1}{2(1 - y)} \]
Thus, the slope of the orthogonal trajectories is:
\[ \frac{dy}{dx} = 2(1 - y) \]

### Step 6: Setting up the differential equation for orthogonal trajectories
We can set up the differential equation for the orthogonal trajectories:
\[ \frac{dy}{dx} = 2(1 - y) \]

### Step 7: Rearranging and separating variables
Rearranging gives us:
\[ \frac{dy}{1 - y} = 2 dx \]

### Step 8: Integrating both sides
Integrating both sides:
\[ \int \frac{dy}{1 - y} = \int 2 dx \]
This results in:
\[ -\ln |1 - y| = 2x + C \]

### Step 9: Exponentiating both sides
Exponentiating both sides gives:
\[ 1 - y = e^{-2x - C} \]
Let \(k = e^{-C}\), we can rewrite this as:
\[ 1 - y = \frac{k}{e^{2x}} \]

### Step 10: Solving for y
Thus, we have:
\[ y = 1 - \frac{k}{e^{2x}} \]

### Final Result
The orthogonal trajectories of the given family of curves are:
\[ y = 1 - ke^{-2x} \]