Question

If the tangent to the curve given by \( x = t^2 - 1 \) and \( y = t^2 - t \) is parallel to the x-axis, then the value of \( t \) is

• A. \( -\frac{1}{\sqrt{3}} \)
• B. 0
• C. \( \frac{1}{\sqrt{3}} \)
• D. \( \frac{1}{2} \)

Hint:

When finding the slope of the tangent line to a curve, differentiate the parametric equations and set \( \frac{dy}{dx} = 0 \) for a horizontal tangent.

Answer 1

The Correct Option is D

Step 1: Find the derivative of \( y \) with respect to \( t \).
We are given \( x = t^2 - 1 \) and \( y = t^2 - t \). To find when the tangent is parallel to the x-axis, we need to find \( \frac{dy}{dx} \). First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): \[ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 2t - 1 \]
Step 2: Set the derivative of \( y \) with respect to \( x \) equal to 0.
The derivative \( \frac{dy}{dx} \) is given by: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t - 1}{2t} \] To have a tangent parallel to the x-axis, \( \frac{dy}{dx} = 0 \). Therefore: \[ \frac{2t - 1}{2t} = 0 \] Solving for \( t \): \[ 2t - 1 = 0 \quad \Rightarrow \quad t = \frac{1}{2} \]
Step 3: Conclusion.
Thus, the value of \( t \) is \( \frac{1}{2} \).