Let a curve `y=f(x),f(x)ge0, AA x in R` has property that for every point P on the curve length of subnormal is equal to abscissa of p. If `f(1)=3`, then f(4) is equal to

Options

A. `-2sqrt6`
B. `2sqrt6`
C. `3sqrt5`
D. None of these

Answer

To solve the problem step by step, we will follow the given conditions and derive the required function.

### Step 1: Understand the given property
We are given that the length of the subnormal at any point \( P \) on the curve \( y = f(x) \) is equal to the abscissa \( x \) of point \( P \). The length of the subnormal can be expressed as:
\[ \text{Length of subnormal} = m \cdot y \]
where \( m \) is the slope of the tangent at point \( P \).

### Step 2: Relate the slope to the function
The slope \( m \) can be expressed as:
\[ m = \frac{dy}{dx} \]
Thus, we can write:
\[ \frac{dy}{dx} \cdot y = x \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ y \frac{dy}{dx} = x \]

### Step 4: Separate variables
We can separate the variables:
\[ y \, dy = x \, dx \]

### Step 5: Integrate both sides
Integrating both sides, we have:
\[ \int y \, dy = \int x \, dx \]
This results in:
\[ \frac{y^2}{2} = \frac{x^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 6: Simplify the equation
Multiplying through by 2 to eliminate the fractions gives:
\[ y^2 = x^2 + 2C \]

### Step 7: Use the initial condition
We are given that \( f(1) = 3 \). Substituting \( x = 1 \) and \( y = 3 \) into the equation:
\[ 3^2 = 1^2 + 2C \]
This simplifies to:
\[ 9 = 1 + 2C \implies 2C = 8 \implies C = 4 \]

### Step 8: Substitute back to find the function
Substituting \( C \) back into the equation gives:
\[ y^2 = x^2 + 8 \]
Thus, we have:
\[ y = \sqrt{x^2 + 8} \]
Since \( f(x) \geq 0 \), we take the positive root.

### Step 9: Find \( f(4) \)
Now, we need to find \( f(4) \):
\[ f(4) = \sqrt{4^2 + 8} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6} \]

### Final Answer
Thus, the value of \( f(4) \) is:
\[ \boxed{2\sqrt{6}} \]