For the curve `x = t^2 - 1, y = t^2 - t`, the tangent line is perpendicular to x-axis, where

Options

A. t=0
B. `t=oo`
C. `t=1//sqrt3`
D. `t=1//sqrt3`

Answer

To find the value of \( t \) for which the tangent line to the curve defined by \( x = t^2 - 1 \) and \( y = t^2 - t \) is perpendicular to the x-axis, we can follow these steps:

### Step 1: Understand the condition for perpendicularity
A line that is perpendicular to the x-axis is a vertical line, which means its slope is undefined. However, in terms of the derivative, we can say that the slope of the tangent line must be zero (horizontal line) for the tangent to be vertical.

### Step 2: Find the derivatives
We need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
- Given \( x = t^2 - 1 \), differentiating with respect to \( t \):
\[ \frac{dx}{dt} = 2t \]
- Given \( y = t^2 - t \), differentiating with respect to \( t \):
\[ \frac{dy}{dt} = 2t - 1 \]

### Step 3: Find the slope of the tangent line
The slope of the tangent line \( \frac{dy}{dx} \) can be found using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t - 1}{2t} \]

### Step 4: Set the slope to zero
For the tangent line to be vertical, we set the slope \( \frac{dy}{dx} \) equal to zero:
\[ \frac{2t - 1}{2t} = 0 \]
This implies:
\[ 2t - 1 = 0 \]

### Step 5: Solve for \( t \)
Solving the equation:
\[ 2t - 1 = 0 \implies 2t = 1 \implies t = \frac{1}{2} \]

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