The tangent at any point `P` of a curve `C` meets the x-axis at `Q` whose abscissa is positive and `OP = OQ, O` being the origin, the equation of curve C satisfying these conditions may be

Options

A. (a)`y^(2) = (1)/(4)(1-4x)`
B. (b)`y^(2) = (1)/(4)(9-12x)`
C. (c)`y^(2) = (1)/(4)(16-16x)`
D. (d)`y^(2) = (1)/(4)(25 - 20 x)`

Answer

To solve the problem, we need to find the equation of the curve \( C \) such that the tangent at any point \( P \) of the curve meets the x-axis at point \( Q \) with the condition that \( OP = OQ \), where \( O \) is the origin.

### Step-by-step Solution:

1. **Define the point \( P \)**:
Let the point \( P \) on the curve \( C \) be represented as \( P(x, y) \).

2. **Equation of the tangent at point \( P \)**:
The equation of the tangent line at point \( P \) can be expressed using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \) at point \( P \). Therefore, the equation becomes:
\[ y - y = \frac{dy}{dx}(x - x) \]

3. **Finding the x-intercept \( Q \)**:
To find the x-intercept \( Q \) where the tangent meets the x-axis, set \( y = 0 \):
\[ 0 - y = \frac{dy}{dx}(x - x) \]
Rearranging gives:
\[ 0 = y - \frac{dy}{dx} (x - x) \]
Thus, the x-coordinate of point \( Q \) is:
\[ Q = x - \frac{y}{\frac{dy}{dx}} \]

4. **Condition \( OP = OQ \)**:
The distance \( OP \) from the origin \( O(0, 0) \) to point \( P(x, y) \) is:
\[ OP = \sqrt{x^2 + y^2} \]
The distance \( OQ \) from the origin \( O(0, 0) \) to point \( Q \) is:
\[ OQ = \sqrt{(x - \frac{y}{\frac{dy}{dx}})^2 + 0^2} = |x - \frac{y}{\frac{dy}{dx}}| \]
Setting these two distances equal gives:
\[ \sqrt{x^2 + y^2} = |x - \frac{y}{\frac{dy}{dx}}| \]

5. **Squaring both sides**:
Squaring both sides to eliminate the square root:
\[ x^2 + y^2 = (x - \frac{y}{\frac{dy}{dx}})^2 \]
Expanding the right-hand side:
\[ x^2 + y^2 = x^2 - 2x\frac{y}{\frac{dy}{dx}} + \left(\frac{y}{\frac{dy}{dx}}\right)^2 \]

6. **Rearranging the equation**:
Cancel \( x^2 \) from both sides:
\[ y^2 = -2x\frac{y}{\frac{dy}{dx}} + \left(\frac{y}{\frac{dy}{dx}}\right)^2 \]

7. **Substituting \( t = \frac{x}{y} \)**:
Let \( t = \frac{x}{y} \), then \( x = yt \) and \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \). Substitute this into the equation.

8. **Integrating**:
After simplifying and integrating, we will arrive at a relation involving \( x \) and \( y \).

9. **Final equation**:
The final equation can be expressed in the form:
\[ y^2 = C - 2x \]
where \( C \) is a constant.

### Conclusion:
The equation of the curve \( C \) satisfying the given conditions is \( y^2 = C - 2x \).