Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1106 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
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
$$
\frac{2 t-1}{2 t}=0 \Rightarrow t=\frac{1}{2}
$$
$$
\begin{array}{l}
\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t} \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-1 \\
\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}}
\end{array}
$$
Since tangent is parallel to $\mathrm{X}$ axis, we write
$$
\frac{2 t-1}{2 t}=0 \Rightarrow t=\frac{1}{2}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.