Q102 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics
If $x^2 \tan ^{-1} \frac{y}{x}-y^2 \tan ^{-1} \frac{x}{y}=k$, then $\left(\frac{d y}{d x}\right)_{(1,1)}=$
- A$0$
- Correct answerB$\pi / 4$
- C$1$
- D$\pi / 2$
Solution
Given, $x^2 \tan ^{-1}\left(\frac{y}{x}\right)-y^2 \tan ^{-1}\left(\frac{x}{y}\right)=K$
Differentiate with respect to $x$, we get.
$\begin{aligned}
& \frac{x^2 d \tan ^{-1}\left(\frac{y}{x}\right)}{d x}+\tan ^{-1}\left(\frac{y}{x}\right) \frac{d\left(x^2\right)}{d x}-\frac{y^2 d \tan ^{-1}\left(\frac{x}{y}\right)}{d x} \\
& -\tan ^{-1}\left(\frac{x}{y}\right) \frac{d\left(y^2\right)}{d x}=0 \\
& \Rightarrow x^2\left(\frac{x^2}{x^2+y^2}\right)\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)+\tan ^{-1}\left(\frac{y}{x}\right) 2 x \\
& -y^2\left(\frac{y^2}{x^2+y^2}\right)\left(\frac{y \cdot \frac{-x d y}{d x}}{y^2}\right)-\tan ^{-1}\left(\frac{x}{y}\right) 2 y=0 \\
& \Rightarrow\left(\frac{1}{2}\right)\left(1 \cdot y_1(1,1)-1\right)+\tan ^{-1}\left(\frac{1}{1}\right) \cdot 2 \\
& -\left(\frac{1}{2}\right)\left(1-y_1(1,1)\right)-\tan ^{-1}\left(\frac{1}{1}\right) \cdot 2=0
\end{aligned}$
$\begin{aligned} & \Rightarrow\left(y_1\left(1_1, 1\right)-1\right)=0 \\ & \Rightarrow y_1(1,1)=1 \Rightarrow \frac{d y}{d x}(1,1)=1\end{aligned}$
Hence, the correct answer is option: 'B'.