Find the point on the curve `y^2=` `8xdot` for which the abscissa and ordinate change at the same rate.

Answer

To find the point on the curve \( y^2 = 8x \) where the abscissa (x-coordinate) and ordinate (y-coordinate) change at the same rate, we follow these steps:

### Step 1: Understand the relationship between rates of change
We are given that the rates of change of x and y with respect to time (t) are equal:
\[ \frac{dx}{dt} = \frac{dy}{dt} \]

### Step 2: Differentiate the curve equation with respect to time
The equation of the curve is:
\[ y^2 = 8x \]
Differentiating both sides with respect to \( t \):
\[ \frac{d}{dt}(y^2) = \frac{d}{dt}(8x) \]
Using the chain rule, we get:
\[ 2y \frac{dy}{dt} = 8 \frac{dx}{dt} \]

### Step 3: Substitute the relationship between rates
Since we know \( \frac{dx}{dt} = \frac{dy}{dt} \), we can substitute \( \frac{dy}{dt} \) for \( \frac{dx}{dt} \):
\[ 2y \frac{dy}{dt} = 8 \frac{dy}{dt} \]

### Step 4: Simplify the equation
Assuming \( \frac{dy}{dt} \neq 0 \) (since we are looking for points where both x and y are changing), we can divide both sides by \( \frac{dy}{dt} \):
\[ 2y = 8 \]

### Step 5: Solve for y
Now, solve for \( y \):
\[ y = \frac{8}{2} = 4 \]

### Step 6: Substitute y back into the curve equation to find x
Now, substitute \( y = 4 \) back into the original curve equation to find \( x \):
\[ (4)^2 = 8x \]
\[ 16 = 8x \]
\[ x = \frac{16}{8} = 2 \]

### Step 7: State the point
Thus, the point on the curve where the abscissa and ordinate change at the same rate is:
\[ (2, 4) \]

### Summary
The point on the curve \( y^2 = 8x \) where the rates of change of x and y are equal is \( (2, 4) \).

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