A particle moves along the cirve `x^(2)+4=y`. The points on the curve at which the y coordinates changes twice as fast as the x coordinate, is

Options

A. `(1, 5)`
B. `(5, 1)`
C. `(1, 2)`
D. None of these

Explanation

To solve the problem, we need to find the points on the curve defined by the equation \( x^2 + 4 = y \) where the rate of change of \( y \) with respect to \( x \) is twice that of \( x \). This means we need to find where \( \frac{dy}{dx} = 2 \).

### Step-by-step Solution:

1. **Identify the curve equation**:
The equation of the curve is given as:
\[ y = x^2 + 4 \]

2. **Differentiate the equation with respect to \( x \)**:
To find the relationship between \( dy \) and \( dx \), we differentiate both sides of the equation:
\[ \frac{dy}{dx} = \frac{d}{dx}(x^2 + 4) \]
The derivative of \( x^2 \) is \( 2x \) and the derivative of a constant (4) is 0. Therefore:
\[ \frac{dy}{dx} = 2x \]

3. **Set up the condition for the rate of change**:
We need to find when the rate of change of \( y \) is twice that of \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{dx}{dx} \]
This simplifies to:
\[ \frac{dy}{dx} = 2 \]

4. **Equate the derivatives**:
Now we equate \( 2x \) to 2:
\[ 2x = 2 \]

5. **Solve for \( x \)**:
Dividing both sides by 2 gives:
\[ x = 1 \]

6. **Find the corresponding \( y \) value**:
Substitute \( x = 1 \) back into the original curve equation to find \( y \):
\[ y = (1)^2 + 4 = 1 + 4 = 5 \]

7. **Conclusion**:
The point on the curve where the \( y \) coordinate changes twice as fast as the \( x \) coordinate is:
\[ (1, 5) \]

### Final Answer:
The point is \( (1, 5) \).

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