If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to $X$ - axis, then the value…
If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to $X$ - axis, then the value of $\mathrm{t}$ is
$\frac{-1}{\sqrt{3}}$
0
$\frac{1}{\sqrt{3}}$
$\frac{1}{2}$
Solution
We have $x=t^{2}-1$ and $y=t^{2}-t$
$\begin{array}{l}
\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t} \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-1 \\
\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}}
\end{array}$
Since tangent is parallel to $\mathrm{X}$ axis, we write
$\frac{2 t-1}{2 t}=0 \Rightarrow t=\frac{1}{2}$