The point on the parabola `y^(2)=4x` at which the abscissa and ordinate change at the same rate is

Options

A. `(2 , 2 sqrt2)`
B. `(2, -2 sqrt2)`
C. `(1, 2)`
D. `(4, 4)`

Answer

To find the point on the parabola \( y^2 = 4x \) where the abscissa (x-coordinate) and ordinate (y-coordinate) change at the same rate, we can follow these steps:

### Step 1: Understand the problem
We need to find a point on the parabola where the rates of change of x and y with respect to time (denoted as \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)) are equal.

### Step 2: Differentiate the equation of the parabola
The equation of the parabola is given by:
\[ y^2 = 4x \]
We differentiate both sides with respect to time \( t \):
\[ \frac{d}{dt}(y^2) = \frac{d}{dt}(4x) \]
Using the chain rule, we get:
\[ 2y \frac{dy}{dt} = 4 \frac{dx}{dt} \]

### Step 3: Set the rates equal
Since we are given that the rates of change are equal, we have:
\[ \frac{dy}{dt} = \frac{dx}{dt} \]
Substituting this into the differentiated equation gives:
\[ 2y \frac{dx}{dt} = 4 \frac{dx}{dt} \]

### Step 4: Simplify the equation
Assuming \( \frac{dx}{dt} \neq 0 \) (since we are looking for a point where the changes are happening), we can divide both sides by \( \frac{dx}{dt} \):
\[ 2y = 4 \]
Thus, we find:
\[ y = 2 \]

### Step 5: Find the corresponding x-coordinate
Now that we have \( y = 2 \), we can substitute this back into the original equation of the parabola to find \( x \):
\[ y^2 = 4x \implies 2^2 = 4x \implies 4 = 4x \implies x = 1 \]

### Step 6: State the point
The point on the parabola where the abscissa and ordinate change at the same rate is:
\[ (1, 2) \]

### Final Answer
The point is \( (1, 2) \).

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