Let C be a curve such that the normal at any point P on it meets x-axis and y-axis at A and B respectively. If BP : PA = 1 : 2 (internally) and the curve passes through the point (0, 4), then which of the following alternative(s) is/are correct?

Options

A. The curves passes through `(sqrt(10), -6)`.
B. The equation of tangent at `(4, 4 sqrt(3)) "is 2x" + sqrt(3) y = 20`.
C. The differential equation for the curve is yy' + 2x = 0.
D. The curve represents a hyperbola.

Explanation

To solve the problem step by step, we will analyze the given information and derive the equation of the curve \( C \).

### Step 1: Understanding the Normal Line
The normal at any point \( P(x, y) \) on the curve meets the x-axis at point \( A \) and the y-axis at point \( B \). The equation of the normal line at point \( P \) can be expressed as:
\[ y - y_1 = -\frac{dy}{dx}(x - x_1) \]
where \( (x_1, y_1) \) is the point \( P \).

### Step 2: Finding Coordinates of Points A and B
1. **Finding Coordinates of A**:
At point \( A \), \( y = 0 \):
\[ 0 - y = -\frac{dy}{dx}(x - x_1) \]
Rearranging gives:
\[ x = x_1 - \frac{y}{\frac{dy}{dx}} \]
Thus, the coordinates of \( A \) are:
\[ A\left(x_1 - \frac{y_1}{\frac{dy}{dx}}, 0\right) \]

2. **Finding Coordinates of B**:
At point \( B \), \( x = 0 \):
\[ y - y_1 = -\frac{dy}{dx}(0 - x_1) \]
Rearranging gives:
\[ y = y_1 + x_1 \frac{dy}{dx} \]
Thus, the coordinates of \( B \) are:
\[ B\left(0, y_1 + x_1 \frac{dy}{dx}\right) \]

### Step 3: Ratio BP : PA
Given that \( BP : PA = 1 : 2 \), we can express this in terms of coordinates:
\[ \frac{BP}{PA} = \frac{y_1 + x_1 \frac{dy}{dx} - y_1}{y_1 - \left(y_1 + x_1 \frac{dy}{dx}\right)} = \frac{x_1 \frac{dy}{dx}}{-x_1 \frac{dy}{dx}} = -1 \]
This implies:
\[ BP = \frac{1}{3} \text{ of the total distance } AB \]

### Step 4: Setting Up the Equation
Using the section formula for the coordinates of \( P \):
\[ P\left(\frac{1 \cdot 0 + 2 \cdot \left(x_1 - \frac{y_1}{\frac{dy}{dx}}\right)}{1 + 2}, \frac{1 \cdot \left(y_1 + x_1 \frac{dy}{dx}\right) + 2 \cdot y_1}{1 + 2}\right) \]
This simplifies to:
\[ P\left(\frac{2x_1 - \frac{2y_1}{\frac{dy}{dx}}}{3}, \frac{3y_1 + x_1 \frac{dy}{dx}}{3}\right) \]

### Step 5: Deriving the Differential Equation
From the above, we can derive:
\[ y \frac{dy}{dx} = 2x \]
This leads to:
\[ y \, dy = 2x \, dx \]

### Step 6: Integrating Both Sides
Integrating both sides gives:
\[ \int y \, dy = \int 2x \, dx \]
This results in:
\[ \frac{y^2}{2} = x^2 + C \]

### Step 7: Applying the Initial Condition
Given that the curve passes through the point \( (0, 4) \):
\[ \frac{4^2}{2} = 0^2 + C \implies 8 = C \]

### Step 8: Final Equation of the Curve
Substituting \( C \) back into the equation:
\[ \frac{y^2}{2} = x^2 + 8 \implies y^2 = 2x^2 + 16 \]

### Conclusion
The equation of the curve is:
\[ y^2 = 2x^2 + 16 \]
This represents a hyperbola.