The tangent at a point P of a curve meets the y-axis at A, and the line parallel to y-axis at A, and the line parallel to y-axis through P meets the x-axis at B. If area of `DeltaOAB` is constant (O being the origin), Then the curve is

Options

A. `cx^(2)-xy+k=0`
B. `y^(2)+2x^(2)=cx`
C. `3x^(2)+4y^(2)=k`
D. `xy-x^(2)y^(2)+kx=0`

Answer

To solve the problem, we need to find the equation of the curve given that the area of triangle OAB is constant, where O is the origin, A is the point where the tangent at point P meets the y-axis, and B is the point where the line parallel to the y-axis at A meets the x-axis.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the coordinates of point P on the curve be \( P(x_1, y_1) \).
- The tangent at point P meets the y-axis at point A, which has coordinates \( A(0, y_1 - m x_1) \), where \( m \) is the slope of the tangent at P.
- The line parallel to the y-axis through point A meets the x-axis at point B, which has coordinates \( B(0, 0) \).

2. **Area of Triangle OAB**:
- The area of triangle OAB can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, the base is the distance from O to B (which is 0) and the height is the distance from O to A (which is \( y_1 - m x_1 \)).
- Therefore, the area can be expressed as:
\[ \text{Area} = \frac{1}{2} \times 0 \times (y_1 - m x_1) = 0 \]
- However, we need to consider the area of triangle OAP, which is:
\[ \text{Area} = \frac{1}{2} \times x_1 \times y_1 \]

3. **Setting the Area to be Constant**:
- Since the area of triangle OAP is constant, we can write:
\[ \frac{1}{2} x_1 y_1 = C \]
- This implies:
\[ x_1 y_1 = 2C \]
- Let's denote \( 2C \) as a constant \( k \), thus:
\[ x_1 y_1 = k \]

4. **Differentiating the Equation**:
- Differentiate both sides with respect to \( x \):
\[ y + x \frac{dy}{dx} = 0 \]
- Rearranging gives:
\[ \frac{dy}{dx} = -\frac{y}{x} \]

5. **Solving the Differential Equation**:
- This is a separable differential equation. We can separate variables:
\[ \frac{dy}{y} = -\frac{dx}{x} \]
- Integrating both sides:
\[ \ln |y| = -\ln |x| + C_1 \]
- Exponentiating gives:
\[ |y| = \frac{C}{|x|} \]
- Thus, we can express this as:
\[ y = \frac{k}{x} \]
- Where \( k \) is a constant.

6. **Conclusion**:
- The equation of the curve is:
\[ y = \frac{k}{x} \]
- This represents a rectangular hyperbola.