The slope of normal at any point P of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point P. Then, the equation of the curve is (where, c is an arbitrary constant)

Options

A. `y^(2)=x+c`
B. `y=ce^(-x^(2))`
C. `y=ce^(-x)`
D. `y^(2)=ln x+c`

Answer

To solve the problem, we need to find the equation of the curve given that the slope of the normal at any point \( P(x, y) \) is the reciprocal of twice the product of the abscissa and the ordinate of point \( P \).

### Step-by-step Solution:

1. **Understanding the Slope of the Normal**:
The slope of the normal at point \( P \) is given as:
\[ m_{\text{normal}} = -\frac{1}{2xy} \]
where \( x \) is the abscissa and \( y \) is the ordinate of point \( P \).

2. **Relating the Slope of the Normal to the Slope of the Tangent**:
The slope of the tangent \( m \) and the slope of the normal \( m_{\text{normal}} \) are related by:
\[ m \cdot m_{\text{normal}} = -1 \]
Therefore, we can express the slope of the tangent as:
\[ m = -\frac{1}{m_{\text{normal}}} = -(-2xy) = 2xy \]
Hence, we have:
\[ \frac{dy}{dx} = 2xy \]

3. **Separating Variables**:
We can rearrange the equation to separate the variables:
\[ \frac{dy}{y} = 2x \, dx \]

4. **Integrating Both Sides**:
Now we integrate both sides:
\[ \int \frac{dy}{y} = \int 2x \, dx \]
The left side integrates to \( \ln |y| \) and the right side integrates to \( x^2 + C \) (where \( C \) is a constant):
\[ \ln |y| = x^2 + C \]

5. **Exponentiating to Solve for \( y \)**:
To solve for \( y \), we exponentiate both sides:
\[ |y| = e^{x^2 + C} = e^{x^2} \cdot e^{C} \]
Let \( e^{C} = c \) (where \( c \) is a positive constant), we have:
\[ y = c e^{x^2} \]

6. **Final Form of the Equation**:
Since we are considering the first quadrant, \( y \) is positive, thus:
\[ y = c e^{x^2} \]

### Conclusion:
The equation of the curve is:
\[ y = c e^{-x^2} \]
where \( c \) is an arbitrary constant.