A family of curves is such that the slope of normal at any point (x, y) is 2(1-y).
If y = f(x) is a member of this family passing through (-1, 2) then its equation is

Options

A. A) `y^(2) - 2y - x-1 = 0`
B. B) `y^(2) + 2y + x - 7 = 0`
C. C) `y^(2) - 2y + x +1 = 0`
D. D) `y^(2) + 2y - x - 9 = 0`

Answer

To solve the problem, we will follow a systematic approach to derive the equation of the family of curves based on the given condition about the slope of the normal.

### Step 1: Understand the relationship between the slope of the normal and the tangent
The slope of the normal at any point \((x, y)\) is given as \(2(1 - y)\). If we denote the slope of the tangent by \(\frac{dy}{dx}\), we can use the relationship:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} \]
Thus, we have:
\[ 2(1 - y) = -\frac{1}{\frac{dy}{dx}} \]
From this, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{1}{2(1 - y)} \]

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

### Step 3: Integrate both sides
Now we will integrate both sides. The left side requires integration with respect to \(y\) and the right side with respect to \(x\):
\[ \int 2(1 - y) dy = \int -dx \]
Calculating the left side:
\[ \int 2(1 - y) dy = 2y - y^2 + C_1 \]
And the right side:
\[ \int -dx = -x + C_2 \]
Equating both sides gives:
\[ 2y - y^2 = -x + C \]
where \(C = C_2 - C_1\).

### Step 4: Rearranging the equation
Rearranging the equation, we have:
\[ y^2 - 2y - x + C = 0 \]

### Step 5: Use the point (-1, 2) to find C
Now we will use the point \((-1, 2)\) to find the constant \(C\):
Substituting \(x = -1\) and \(y = 2\):
\[ 2^2 - 2(2) - (-1) + C = 0 \]
This simplifies to:
\[ 4 - 4 + 1 + C = 0 \implies C = -1 \]

### Step 6: Substitute C back into the equation
Substituting \(C = -1\) back into the equation:
\[ y^2 - 2y - x - 1 = 0 \]
This can be rearranged to:
\[ y^2 - 2y - x - 1 = 0 \]

### Final Answer
The equation of the family of curves is:
\[ y^2 - 2y - x - 1 = 0 \]