Let y=f(x)be curve passing through `(1,sqrt3)` such that tangent at any point P on the curve lying in the first quadrant has positive slope and the tangent and the normal at the point P cut the x-axis at A and B respectively so that the mid-point of AB is origin. Find the differential equation of the curve and hence determine f(x).

Answer

To solve the problem step by step, we need to find the differential equation of the curve \( y = f(x) \) that passes through the point \( (1, \sqrt{3}) \) and meets the given conditions regarding the tangent and normal lines.

### Step 1: Define the point and slopes
Let \( P(x, f(x)) \) be a point on the curve in the first quadrant. The slope of the tangent at point \( P \) is given by \( f'(x) \), which is positive since the tangent has a positive slope.

### Step 2: Write the equations of the tangent and normal
The equation of the tangent line at point \( P \) is:
\[ y - f(x) = f'(x)(x - x) \]
This simplifies to:
\[ y = f'(x)(x - x) + f(x) \]

The equation of the normal line at point \( P \) is:
\[ y - f(x) = -\frac{1}{f'(x)}(x - x) \]
This simplifies to:
\[ y = -\frac{1}{f'(x)}(x - x) + f(x) \]

### Step 3: Find the x-intercepts of the tangent and normal
Let \( A \) be the x-intercept of the tangent line and \( B \) be the x-intercept of the normal line.

For the tangent line:
Setting \( y = 0 \):
\[ 0 = f'(x)(x - A) + f(x) \implies A = x - \frac{f(x)}{f'(x)} \]

For the normal line:
Setting \( y = 0 \):
\[ 0 = -\frac{1}{f'(x)}(x - B) + f(x) \implies B = x + f(x)f'(x) \]

### Step 4: Midpoint condition
The midpoint of \( AB \) is given by:
\[ \left( \frac{A + B}{2}, 0 \right) = (0, 0) \]
This implies:
\[ \frac{A + B}{2} = 0 \implies A + B = 0 \implies A = -B \]

### Step 5: Set up the equation
Substituting the expressions for \( A \) and \( B \):
\[ x - \frac{f(x)}{f'(x)} + x + f(x)f'(x) = 0 \]
This simplifies to:
\[ 2x + f(x)f'(x) - \frac{f(x)}{f'(x)} = 0 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ f(x)f'(x) = \frac{f(x)}{f'(x)} - 2x \]
Multiplying through by \( f'(x) \):
\[ f(x)(f'(x))^2 = f(x) - 2xf'(x) \]

### Step 7: Form the differential equation
This leads us to the differential equation:
\[ f(x)(f'(x))^2 + 2xf'(x) - f(x) = 0 \]

### Step 8: Solve the differential equation
To solve this, we can separate variables or use an integrating factor. However, we can also look for a solution directly.

Assuming a solution of the form \( f(x) = \sqrt{1 + 2x} \), we can check if it satisfies the original conditions.

### Step 9: Verify the solution
Substituting \( f(x) = \sqrt{1 + 2x} \):
1. Check that it passes through \( (1, \sqrt{3}) \):
\[ f(1) = \sqrt{1 + 2 \cdot 1} = \sqrt{3} \]
This is correct.

2. Check the slope:
\[ f'(x) = \frac{1}{2\sqrt{1 + 2x}} \cdot 2 = \frac{1}{\sqrt{1 + 2x}} \]
This is positive for \( x > 0 \).

Thus, the function \( f(x) = \sqrt{1 + 2x} \) satisfies all conditions.

### Final Answer:
The function is:
\[ f(x) = \sqrt{1 + 2x} \]