If the tangent to the curve `x^(3)-y^(2) = 0` at `(m^(2), -m^(2))` is parallel to `y= -(1)/(m) x-2m^(3)`, then the value of `m^(3)` is

Options

A. `(1//3)`
B. `1//6`
C. `2//3`
D. `-2//3`

Answer

To solve the problem step by step, we need to find the value of \( m^3 \) given that the tangent to the curve \( x^3 - y^2 = 0 \) at the point \( (m^2, -m^2) \) is parallel to the line \( y = -\frac{1}{m}x - 2m^3 \).

### Step 1: Differentiate the curve
We start with the equation of the curve:
\[ x^3 - y^2 = 0 \]
To find the slope of the tangent, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^3) - \frac{d}{dx}(y^2) = 0 \]
Using the chain rule, we get:
\[ 3x^2 - 2y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ 2y \frac{dy}{dx} = 3x^2 \]
Thus,
\[ \frac{dy}{dx} = \frac{3x^2}{2y} \]

### Step 3: Substitute the point \((m^2, -m^2)\)
Now we substitute \( x = m^2 \) and \( y = -m^2 \) into the derivative:
\[ \frac{dy}{dx} = \frac{3(m^2)^2}{2(-m^2)} = \frac{3m^4}{-2m^2} = -\frac{3m^2}{2} \]

### Step 4: Find the slope of the given line
The slope of the line \( y = -\frac{1}{m}x - 2m^3 \) is \( -\frac{1}{m} \).

### Step 5: Set the slopes equal
Since the tangent to the curve is parallel to the line, we set the slopes equal:
\[ -\frac{3m^2}{2} = -\frac{1}{m} \]
Removing the negative signs gives:
\[ \frac{3m^2}{2} = \frac{1}{m} \]

### Step 6: Cross-multiply to solve for \( m \)
Cross-multiplying yields:
\[ 3m^2 \cdot m = 2 \implies 3m^3 = 2 \]

### Step 7: Solve for \( m^3 \)
Dividing both sides by 3 gives:
\[ m^3 = \frac{2}{3} \]

### Final Answer
Thus, the value of \( m^3 \) is:
\[ \boxed{\frac{2}{3}} \]