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:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} \]
If we denote the slope of the tangent as \( \frac{dy}{dx} \), then:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} \]

2. **Setting Up the Equation**:
According to the problem, the slope of the normal is also given as the reciprocal of twice the product of the abscissa and the ordinate:
\[ \text{slope of normal} = -\frac{1}{2xy} \]
Therefore, we can equate the two expressions for the slope of the normal:
\[ -\frac{1}{\frac{dy}{dx}} = -\frac{1}{2xy} \]

3. **Finding the Slope of the Tangent**:
By simplifying the equation, we have:
\[ \frac{1}{\frac{dy}{dx}} = \frac{1}{2xy} \]
Taking the reciprocal of both sides gives us:
\[ \frac{dy}{dx} = -2xy \]

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

5. **Integrating Both Sides**:
Now we integrate both sides:
\[ \int \frac{dy}{y} = \int -2x \, dx \]
This results in:
\[ \ln |y| = -x^2 + C \]
where \( C \) is the constant of integration.

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

7. **Final Equation of the Curve**:
Thus, the equation of the curve is:
\[ y = c e^{-x^2} \]
where \( c \) is an arbitrary constant.