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KCET · Maths · Differential Equations

If \(\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right)\), then

  1. A \(x y_{1}=n \sqrt{b^{2}-y^{2}}\)
  2. B \(x y_{1}+n \sqrt{b^{2}-y^{2}}=0\)
  3. C \(y_{1}=x \sqrt{b^{2}-y^{2}}\)
  4. D \(x y_{1}-\sqrt{b^{2}-y^{2}}=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(x y_{1}+n \sqrt{b^{2}-y^{2}}=0\)

Step-by-step Solution

Detailed explanation

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. ' \(x\) '
\[
\begin{gathered}
-\frac{1}{\sqrt{1-(y / b)^{2}}} \cdot \frac{y_{1}}{b}=n \cdot \frac{1}{(x / n)} \cdot \frac{1}{n} \\
-\frac{b}{\sqrt{b^{2}-y^{2}}} \cdot \frac{y_{1}}{b}=\frac{n^{2}}{x} \cdot \frac{1}{n} \\
\Rightarrow \quad-\frac{y_{1}}{\sqrt{b^{2}-y^{2}}}=\frac{n}{x} \\
\Rightarrow \quad-x y_{1}=n \sqrt{b^{2}-y^{2}} \\
\Rightarrow \quad x y_{1}+n \sqrt{b^{2}-y^{2}}=0
\end{gathered}
\]