The slope of the tangent to the curve x = 3t2 + 1, y = t3 – 1 at x = 1 is

Q17 of 149 Page 16

The slope of the tangent to the curve x = 3t² + 1, y = t³ – 1 at x = 1 is

Given that x = 3t² + 1, y = t³ – 1

For x = 1,

3t² + 1=1

⇒ 3t² = 0

⇒ t =0

Now, differentiating both the equations w.r.t. t, we get

⇒Slope of the curve:

For t =0,

Slope of the curve =0

Hence, option B is correct.

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