A particle moves along the parabola `y^(2)=4x`. Find the co - ordinates of the point on the parabola where the rate of increment of abscissa is twice the rate of increment of the ordinate.

Answer

To solve the problem, we need to find the coordinates of the point on the parabola \( y^2 = 4x \) where the rate of increment of the abscissa (x-coordinate) is twice the rate of increment of the ordinate (y-coordinate).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have a parabola defined by the equation \( y^2 = 4x \).
- We need to find the point where the rate of change of x (abscissa) with respect to time \( \frac{dx}{dt} \) is twice the rate of change of y (ordinate) with respect to time \( \frac{dy}{dt} \). This can be expressed as:
\[ \frac{dx}{dt} = 2 \frac{dy}{dt} \]

2. **Differentiate the Parabola with Respect to Time**:
- We differentiate the equation \( y^2 = 4x \) with respect to time \( t \):
\[ \frac{d}{dt}(y^2) = \frac{d}{dt}(4x) \]
- Applying the chain rule, we get:
\[ 2y \frac{dy}{dt} = 4 \frac{dx}{dt} \]

3. **Substituting the Rate of Change Relationship**:
- From step 1, we know \( \frac{dx}{dt} = 2 \frac{dy}{dt} \). We substitute this into the differentiated equation:
\[ 2y \frac{dy}{dt} = 4(2 \frac{dy}{dt}) \]
- Simplifying this gives:
\[ 2y \frac{dy}{dt} = 8 \frac{dy}{dt} \]

4. **Isolating \( \frac{dy}{dt} \)**:
- Assuming \( \frac{dy}{dt} \neq 0 \) (since it represents a change), we can divide both sides by \( \frac{dy}{dt} \):
\[ 2y = 8 \]
- Solving for \( y \):
\[ y = 4 \]

5. **Finding the Corresponding x-coordinate**:
- Now that we have \( y = 4 \), we can find \( x \) using the original parabola equation:
\[ y^2 = 4x \implies 4^2 = 4x \implies 16 = 4x \]
- Solving for \( x \):
\[ x = \frac{16}{4} = 4 \]

6. **Final Coordinates**:
- The coordinates of the point on the parabola where the rate of increment of the abscissa is twice the rate of increment of the ordinate are:
\[ (x, y) = (4, 4) \]

### Final Answer:
The coordinates of the point are \( (4, 4) \).