Search any question & find its solution
Question:
Answered & Verified by Expert
$x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 t}{1+t^2}$, then $\frac{d y}{d x}=$
Options:
Solution:
1580 Upvotes
Verified Answer
The correct answer is:
$\frac{-x}{y}$
$x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 t}{1+t^2}$
Put $t=\tan \theta$ in both the equations, we get
$x=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\cos 2 \theta$ and $y=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta$.
Differentiating both the equations, we get
$\frac{d x}{d \theta}=-2 \sin 2 \theta$ and $\frac{d y}{d \theta}=2 \cos 2 \theta$.
Therefore $\frac{d y}{d x}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$.
Put $t=\tan \theta$ in both the equations, we get
$x=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\cos 2 \theta$ and $y=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta$.
Differentiating both the equations, we get
$\frac{d x}{d \theta}=-2 \sin 2 \theta$ and $\frac{d y}{d \theta}=2 \cos 2 \theta$.
Therefore $\frac{d y}{d x}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$.
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.