The normal at a general point (a, b) on curve makes an angle `theta` with x-axis which satisfies `b(a^(2)tan theta - cot theta)=a(b^(2)+1)`. The equation of curve can be

Options

A. `y = e^(x^(2)//2)+c`
B. `log ky^(2) = x^(2)`
C. `y = ke^(x^(2)//2)`
D. `x^(2)-y^(2)=k`

Explanation

To find the equation of the curve given the condition on the normal, we will follow these steps:

### Step 1: Understand the slope of the normal
The slope of the normal at a point (a, b) on the curve is given by:
\[ \text{slope of normal} = -\frac{dx}{dy} \]
This means that if we denote the slope of the tangent by \( \frac{dy}{dx} \), we have:
\[ \frac{dy}{dx} = -\frac{1}{\text{slope of normal}} \]

### Step 2: Rewrite the given equation
The given equation is:
\[ b(a^2 \tan \theta - \cot \theta) = a(b^2 + 1) \]
Substituting \( a = x \), \( b = y \), \( \tan \theta = -\frac{dx}{dy} \), and \( \cot \theta = \frac{dy}{dx} \), we can rewrite the equation as:
\[ y(x^2(-\frac{dx}{dy}) - \frac{dy}{dx}) = x(y^2 + 1) \]

### Step 3: Simplify the equation
Rearranging gives:
\[ -y x^2 \frac{dx}{dy} + y \frac{dy}{dx} = x(y^2 + 1) \]
Now, we can multiply through by \( dy \):
\[ -y x^2 dx + y dy = x(y^2 + 1) dy \]
This leads to:
\[ -y x^2 dx + y dy - x(y^2 + 1) dy = 0 \]

### Step 4: Rearranging terms
Rearranging the terms gives:
\[ y dy - y x^2 dx - x(y^2 + 1) dy = 0 \]
This can be rearranged to:
\[ y dy + (-y x^2 - x(y^2 + 1)) dy = 0 \]

### Step 5: Factor the equation
Factoring out \( dy \):
\[ dy \left( y - x(y^2 + 1) - y x^2 \right) = 0 \]
This gives us:
\[ y dy - (x y^2 + x) dy - y x^2 = 0 \]

### Step 6: Solve for dy/dx
Setting the factor equal to zero:
\[ y dy - x(y^2 + 1) dy - y x^2 = 0 \]
This can be rearranged to:
\[ y \frac{dy}{dx} - x(y^2 + 1) = 0 \]
Thus:
\[ \frac{dy}{dx} = \frac{x(y^2 + 1)}{y} \]

### Step 7: Integrate to find the curve
Now we can separate variables and integrate:
\[ \frac{y}{y^2 + 1} dy = x dx \]
Integrating both sides:
\[ \int \frac{y}{y^2 + 1} dy = \int x dx \]
The left side integrates to:
\[ \frac{1}{2} \ln(y^2 + 1) = \frac{x^2}{2} + C \]

### Step 8: Final equation of the curve
Exponentiating both sides gives:
\[ y^2 + 1 = k e^{x^2} \]
where \( k = e^{2C} \). Thus, the equation of the curve can be expressed as:
\[ y^2 = k e^{x^2} - 1 \]