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If $\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right)$, then
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Verified Answer
The correct answer is:
$x y_{1}+n \sqrt{b^{2}-y^{2}}=0$
Given, $\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right) \quad\left(\because y_{1}=\frac{d y}{d x}\right)$
Differentiating w.r.t. ' $x$ '
$$
\begin{gathered}
-\frac{1}{\sqrt{1-(y / b)^{2}}} \cdot \frac{y_{1}}{b}=n \cdot \frac{1}{(x / n)} \cdot \frac{1}{n} \\
-\frac{b}{\sqrt{b^{2}-y^{2}}} \cdot \frac{y_{1}}{b}=\frac{n^{2}}{x} \cdot \frac{1}{n} \\
\Rightarrow \quad-\frac{y_{1}}{\sqrt{b^{2}-y^{2}}}=\frac{n}{x} \\
\Rightarrow \quad-x y_{1}=n \sqrt{b^{2}-y^{2}} \\
\Rightarrow \quad x y_{1}+n \sqrt{b^{2}-y^{2}}=0
\end{gathered}
$$
Differentiating w.r.t. ' $x$ '
$$
\begin{gathered}
-\frac{1}{\sqrt{1-(y / b)^{2}}} \cdot \frac{y_{1}}{b}=n \cdot \frac{1}{(x / n)} \cdot \frac{1}{n} \\
-\frac{b}{\sqrt{b^{2}-y^{2}}} \cdot \frac{y_{1}}{b}=\frac{n^{2}}{x} \cdot \frac{1}{n} \\
\Rightarrow \quad-\frac{y_{1}}{\sqrt{b^{2}-y^{2}}}=\frac{n}{x} \\
\Rightarrow \quad-x y_{1}=n \sqrt{b^{2}-y^{2}} \\
\Rightarrow \quad x y_{1}+n \sqrt{b^{2}-y^{2}}=0
\end{gathered}
$$
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