Q57 / 125 · BITSAT 2021 · Mathematics
If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$, then $\frac{d y}{d x}$ is equal to :
- A$-\frac{y}{x}$
- B$\frac{y}{x}$
- Correct answerC$-\frac{x}{y}$
- D$\frac{x}{y}$
Solution
Let $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$
Put $t=\tan \theta$, we get
$x=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ and $y=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$
$\Rightarrow x=\cos 2 \theta$ and $y=\sin 2 \theta$
$\therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=-2 \sin 2 \theta$ and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=2 \cos 2 \theta$
Now, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{dx}}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$
Hence, the correct answer is option: 'C'.