Answer
\[(4,0)\]
Work Step by Step
\[x=3t^2+1\;,\;y=t^3-1\]
Differentiate $x$ with respect to $t$:
\[\frac{dx}{dt}=6t\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{dt}=3t^2\]
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^2}{6t}\]
\[\Rightarrow \frac{dy}{dx}=\frac{1}{2}t\;\;\;\;\;\ldots (1)\]
The slope of tangent line at ($a,b$) is given by $\displaystyle\left.\frac{dy}{dx}\right|_{(a,b)}$
Given that slope that slope is equal to $\displaystyle\frac{1}{2}$
Using (1)
\[\Rightarrow \frac{1}{2}t=\frac{1}{2}\Rightarrow t=1\]
\[\Rightarrow x=3(1)^2+1=4\]
\[\Rightarrow y=(1)^3-1=0\]
Therefore the required point is $\displaystyle\left(4,0\right)$.
