Let y= f(x) be a curve which passes through (3,1) and is such that normal at any point on it passes through (1,1). Then y= f(x) describes

Options

A. a circle of area `pi`
B. an ellipse of area `2pi`
C. an ellipse of area `3pi`
D. a circle of area `4pi`

Answer

To solve the problem, we need to determine the equation of the curve \( y = f(x) \) given that the normal at any point on the curve passes through the point (1, 1) and that the curve passes through the point (3, 1).

### Step 1: Understanding the Normal Line
The normal line to the curve at any point \( (x, f(x)) \) has a slope that is the negative reciprocal of the derivative of the curve at that point. If \( f'(x) \) is the derivative of \( f(x) \), then the slope of the normal line is \( -\frac{1}{f'(x)} \).

### Step 2: Equation of the Normal Line
The equation of the normal line at the point \( (x, f(x)) \) can be expressed as:
\[ y - f(x) = -\frac{1}{f'(x)}(x - x_0) \]
Where \( (x_0, f(x_0)) \) is the point on the curve. Rearranging gives:
\[ y = -\frac{1}{f'(x)}(x - x_0) + f(x) \]

### Step 3: Condition for the Normal Line
Since the normal line passes through the point (1, 1), we can substitute \( (1, 1) \) into the normal line equation:
\[ 1 = -\frac{1}{f'(x)}(1 - x) + f(x) \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ f(x) = 1 + \frac{1 - x}{f'(x)} \]
This implies:
\[ f'(x) f(x) = f'(x) + 1 - x \]

### Step 5: Rearranging Further
Rearranging gives a differential equation:
\[ f'(x) f(x) - f'(x) = 1 - x \]
Factoring out \( f'(x) \):
\[ f'(x)(f(x) - 1) = 1 - x \]

### Step 6: Solving the Differential Equation
This is a separable differential equation. We can separate the variables:
\[ \frac{f'(x)}{1 - x} = \frac{1}{f(x) - 1} \]

### Step 7: Integrating Both Sides
Integrating both sides gives:
\[ \int \frac{1}{1 - x} dx = \int \frac{1}{f(x) - 1} df \]
This results in:
\[ -\ln|1 - x| = \ln|f(x) - 1| + C \]

### Step 8: Exponentiating
Exponentiating both sides gives:
\[ |f(x) - 1| = \frac{K}{|1 - x|} \]
Where \( K = e^{-C} \).

### Step 9: Finding the Constant
Using the point (3, 1) to find \( K \):
\[ |f(3) - 1| = \frac{K}{|1 - 3|} \implies |1 - 1| = \frac{K}{2} \implies K = 0 \]
Thus, \( f(x) = 1 \).

### Conclusion
The curve described by \( y = f(x) \) is a horizontal line at \( y = 1 \).

### Final Answer
The equation of the curve is:
\[ y = 1 \]