The slope of the tangent to the curve `x=3t^2+1` , `y=t^3-1` at `x=1` is `1//2` (b) 0 (c) -2 (d) `oo`

Explanation

Given, `x=3t^2+1,y=t^3−1`

Slope of the tangent to the given curve is `dy/dx​=(dy/dt)​×(dt/dx)`
​
=`3t^2×1/6t`
​
=2t​

Since the slope has to be calculated at x=1,

i.e. at` 3t^2+1=1,`

we get t=0

Thus, the required slope is 0.