On the curve `y= x^(3)`, the point at which the tangent line is parallel to the chord through the point `(-1,-1)` and (2,8) is

Options

A. (1,1)
B. `((1)/(2), (1)/(8))`
C. `((1)/(3), (1)/(27))`
D. `((1)/(2), -(1)/(8))`

Answer

To find the point on the curve \( y = x^3 \) where the tangent line is parallel to the chord connecting the points \((-1, -1)\) and \((2, 8)\), we will follow these steps:

### Step 1: Find the slope of the chord
The slope of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
For the points \((-1, -1)\) and \((2, 8)\):
\[ \text{slope} = \frac{8 - (-1)}{2 - (-1)} = \frac{8 + 1}{2 + 1} = \frac{9}{3} = 3 \]

### Step 2: Find the derivative of the curve
The curve is given by \( y = x^3 \). To find the slope of the tangent line at any point on the curve, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3x^2 \]

### Step 3: Set the derivative equal to the slope of the chord
We want the slope of the tangent line to be equal to the slope of the chord, which we found to be \( 3 \):
\[ 3x^2 = 3 \]

### Step 4: Solve for \( x \)
Dividing both sides by 3:
\[ x^2 = 1 \]
Taking the square root of both sides gives:
\[ x = 1 \quad \text{or} \quad x = -1 \]

### Step 5: Find the corresponding \( y \) values
Now we will find the \( y \) values for both \( x \) values using the equation of the curve \( y = x^3 \):
1. For \( x = 1 \):
\[ y = 1^3 = 1 \quad \Rightarrow \quad (1, 1) \]
2. For \( x = -1 \):
\[ y = (-1)^3 = -1 \quad \Rightarrow \quad (-1, -1) \]

### Step 6: Identify the valid point
Since the point \((-1, -1)\) is already given in the problem, we discard it. Thus, the point on the curve where the tangent line is parallel to the chord is:
\[ \boxed{(1, 1)} \]