A family of curves is such that the slope of normal at any point (x, y) is 2(1-y).
The area bounded by the curve y = f(x) of question number 1 and the line x+2y = 0 is

Options

A. `(10)/(3)` sq. units
B. `(4)/(3)` sq. units
C. `(28)/(3)` sq. units
D. `(16)/(3)` sq. units

Answer

To solve the problem step by step, we need to find the family of curves defined by the given slope of the normal and then calculate the area bounded by this curve and the line \( x + 2y = 0 \).

### Step 1: Understanding the slope of the normal
The problem states that the slope of the normal at any point \((x, y)\) is given by:
\[ \text{slope of normal} = 2(1 - y) \]

### Step 2: Finding the slope of the tangent
The slope of the tangent can be found using the relationship:
\[ \text{slope of tangent} = -\frac{1}{\text{slope of normal}} = -\frac{1}{2(1 - y)} \]
Thus,
\[ \frac{dy}{dx} = -\frac{1}{2(1 - y)} \]

### Step 3: Separating variables
We can rewrite the equation to separate variables:
\[ (1 - y) dy = -\frac{1}{2} dx \]

### Step 4: Integrating both sides
Now we integrate both sides:
\[ \int (1 - y) dy = -\frac{1}{2} \int dx \]
This gives us:
\[ y - \frac{y^2}{2} = -\frac{1}{2} x + C \]
Rearranging, we have:
\[ y - \frac{y^2}{2} + \frac{1}{2} x = C \]

### Step 5: Rearranging the equation
To express this in a more standard form, we can multiply through by 2:
\[ 2y - y^2 + x = 2C \]
Let \(C' = 2C\), then:
\[ y^2 - 2y + x = C' \]

### Step 6: Finding the line equation
The line given is:
\[ x + 2y = 0 \quad \Rightarrow \quad y = -\frac{1}{2}x \]

### Step 7: Finding points of intersection
To find the area bounded by the curve and the line, we need to find the points of intersection. Substitute \(y = -\frac{1}{2}x\) into the curve equation:
\[ (-\frac{1}{2}x)^2 - 2(-\frac{1}{2}x) + x = C' \]
This simplifies to:
\[ \frac{1}{4}x^2 + x + x = C' \quad \Rightarrow \quad \frac{1}{4}x^2 + 2x - C' = 0 \]

### Step 8: Solving the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot \frac{1}{4} \cdot (-C')}}{2 \cdot \frac{1}{4}} = \frac{-2 \pm \sqrt{4 + C'}}{\frac{1}{2}} = -4 \pm 2\sqrt{4 + C'} \]
Let the roots be \(x_1\) and \(x_2\).

### Step 9: Finding the area
The area \(A\) between the curve and the line from \(x_1\) to \(x_2\) is given by:
\[ A = \int_{x_1}^{x_2} \left( f(x) - (-\frac{1}{2}x) \right) dx \]
Where \(f(x)\) is the expression for \(y\) derived from the curve.

### Step 10: Conclusion
The exact area can be computed once the specific value of \(C'\) is known, as it determines the specific curve. Without loss of generality, we can denote the area as:
\[ A = \int_{x_1}^{x_2} \left( f(x) + \frac{1}{2}x \right) dx \]