Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{x}=\mathrm{a}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)$ and $\mathrm{y}=\mathrm{b}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}}=$
Options:
Solution:
1389 Upvotes
Verified Answer
The correct answer is:
$\frac{b^2 x}{a^2 y}$
$$
\begin{aligned}
& \mathrm{x}=\mathrm{a}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right) \text { and } \mathrm{y}=\mathrm{b}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right) \\
& \therefore \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}\left(1+\frac{1}{\mathrm{t}^2}\right) \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}\left(1-\frac{1}{\mathrm{t}^2}\right) \\
& \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b}\left(1-\frac{1}{\mathrm{t}^2}\right)}{\mathrm{a}\left(1+\frac{1}{\mathrm{t}^2}\right)}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{t}^2-1}{\mathrm{t}^2+1}\right)
\end{aligned}
$$
Now $\mathrm{x}=\mathrm{a}\left(\frac{\mathrm{t}^2-1}{\mathrm{t}}\right)$ and $\mathrm{y}=\mathrm{b}\left(\frac{\mathrm{t}^2+1}{\mathrm{t}}\right)$
...[From (1)]
$$
\therefore\left(\mathrm{t}^2-1\right)=\left(\frac{\mathrm{x}}{\mathrm{a}}\right) \mathrm{t} \text { and }\left(\mathrm{t}^2+1\right)=\left(\frac{\mathrm{y}}{\mathrm{b}}\right)(\mathrm{t})
$$
$\therefore$ Eq. (2) becomes
$$
\frac{d y}{d x}=\left(\frac{b}{a}\right)\left(\frac{x}{a}\right)(t) \times \frac{1}{\left(\frac{y}{b}\right) \times t}=\frac{b^2 x}{a^2 y}
$$
\begin{aligned}
& \mathrm{x}=\mathrm{a}\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right) \text { and } \mathrm{y}=\mathrm{b}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right) \\
& \therefore \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}\left(1+\frac{1}{\mathrm{t}^2}\right) \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}\left(1-\frac{1}{\mathrm{t}^2}\right) \\
& \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b}\left(1-\frac{1}{\mathrm{t}^2}\right)}{\mathrm{a}\left(1+\frac{1}{\mathrm{t}^2}\right)}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{t}^2-1}{\mathrm{t}^2+1}\right)
\end{aligned}
$$
Now $\mathrm{x}=\mathrm{a}\left(\frac{\mathrm{t}^2-1}{\mathrm{t}}\right)$ and $\mathrm{y}=\mathrm{b}\left(\frac{\mathrm{t}^2+1}{\mathrm{t}}\right)$
...[From (1)]
$$
\therefore\left(\mathrm{t}^2-1\right)=\left(\frac{\mathrm{x}}{\mathrm{a}}\right) \mathrm{t} \text { and }\left(\mathrm{t}^2+1\right)=\left(\frac{\mathrm{y}}{\mathrm{b}}\right)(\mathrm{t})
$$
$\therefore$ Eq. (2) becomes
$$
\frac{d y}{d x}=\left(\frac{b}{a}\right)\left(\frac{x}{a}\right)(t) \times \frac{1}{\left(\frac{y}{b}\right) \times t}=\frac{b^2 x}{a^2 y}
$$
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.