If the tangent to the curve `x = t^(2) - 1, y = t^(2) - t` is parallel to x-axis , then

Options

A. t = 0
B. t = 2
C. ` t = (1)/(2)`
D. `t = - (1)/(2)`

Answer

To solve the problem, we need to find the values of \( t \) for which the tangent to the curve defined by the parametric equations \( x = t^2 - 1 \) and \( y = t^2 - t \) is parallel to the x-axis. A tangent line is parallel to the x-axis when its slope is zero, which means we need to find when \( \frac{dy}{dx} = 0 \).

### Step-by-Step Solution:

1. **Find \( \frac{dy}{dx} \)**:
We will use the chain rule to find \( \frac{dy}{dx} \) in terms of \( t \):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

2. **Differentiate \( y \) with respect to \( t \)**:
Given \( y = t^2 - t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(t^2 - t) = 2t - 1 \]

3. **Differentiate \( x \) with respect to \( t \)**:
Given \( x = t^2 - 1 \):
\[ \frac{dx}{dt} = \frac{d}{dt}(t^2 - 1) = 2t \]

4. **Substitute into \( \frac{dy}{dx} \)**:
Now substituting \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) into the formula for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2t - 1}{2t} \]

5. **Set \( \frac{dy}{dx} = 0 \)**:
For the tangent to be parallel to the x-axis, we set \( \frac{dy}{dx} = 0 \):
\[ \frac{2t - 1}{2t} = 0 \]

6. **Solve for \( t \)**:
The numerator must be zero for the fraction to be zero:
\[ 2t - 1 = 0 \]
\[ 2t = 1 \implies t = \frac{1}{2} \]

7. **Check for other possible values**:
Since the problem asks for all values of \( t \) where the tangent is parallel to the x-axis, we also need to consider the cases where \( t \) could be negative or positive. However, based on the derivative we calculated, the only solution from the equation \( 2t - 1 = 0 \) is \( t = \frac{1}{2} \).

### Conclusion:
The value of \( t \) for which the tangent to the curve is parallel to the x-axis is:
\[ t = \frac{1}{2} \]