A curve in the first quadrant is such that the slope of OP is twice the slope of the tangent drawn at P to the curve, where O is the origin and P is any general point on the curve. If the curve passes through (4, 2), then its equation is

Options

A. `y=x^(2)-14`
B. `y^(2)=x`
C. `y=x^(3)-62`
D. `y=sin(x-4)+2`

Answer

To solve the problem, we need to find the equation of the curve in the first quadrant such that the slope of the line connecting the origin (O) to any point P on the curve is twice the slope of the tangent to the curve at that point P. The curve passes through the point (4, 2).

### Step-by-step Solution:

1. **Understanding the Slopes**:
- Let \( P(x, y) \) be a point on the curve.
- The slope of the line \( OP \) (from the origin to point P) is given by:
\[ \text{slope of } OP = \frac{y}{x} \]
- Let the slope of the tangent to the curve at point P be \( \frac{dy}{dx} \).

2. **Setting up the Relationship**:
- According to the problem, the slope of \( OP \) is twice the slope of the tangent at point P:
\[ \frac{y}{x} = 2 \cdot \frac{dy}{dx} \]

3. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{dy}{dx} = \frac{y}{2x} \]

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

5. **Integrating Both Sides**:
- Integrating both sides:
\[ \int \frac{2 \, dy}{y} = \int \frac{dx}{x} \]
- This gives:
\[ 2 \ln |y| = \ln |x| + C \]

6. **Exponentiating**:
- Exponentiating both sides results in:
\[ |y|^2 = e^{C} |x| \]
- Let \( e^{C} = k \) (a constant), so:
\[ y^2 = kx \]

7. **Using the Given Point to Find k**:
- The curve passes through the point (4, 2):
\[ 2^2 = k \cdot 4 \]
\[ 4 = 4k \implies k = 1 \]

8. **Final Equation of the Curve**:
- Substituting \( k \) back into the equation gives:
\[ y^2 = x \]

### Conclusion:
The equation of the curve is:
\[ \boxed{y^2 = x} \]