Let cos ^{-1}left(frac{y}{b}right)=log left(f | vedclass

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(A) Given: $\cos ^{-1}\left(\frac{y}{b}\right) = n \log \left(\frac{x}{n}\right)$.Taking the derivative with respect to $x$:$-\frac{1}{\sqrt{1 - (y/b)

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$x^{2} y_{2}+x y_{1}+n^{2} y=0$

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(A) Given: $\cos ^{-1}\left(\frac{y}{b}ight) = n \log \left(\frac{x}{n}ight)$.Taking the derivative with respect to $x$:$-\frac{1}{\sqrt{1 - (y/b)^2}} \cdot \frac{y_1}{b} = n \cdot \frac{n}{x} \cdot \frac{1}{n} = \frac{n}{x}$.$-\frac{1}{\sqrt{(b^2 - y^2)/b^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$ $\Rightarrow -\frac{b}{\sqrt{b^2 - y^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$.$-\frac{y_1}{\sqrt{b^2 - y^2}} = \frac{n}{x} \Rightarrow -x y_1 = n \sqrt{b^2 - y^2}$.Squaring both sides: $x^2 y_1^2 = n^2 (b^2 - y^2)$.Differentiating again with respect to $x$:$x^2 (2 y_1 y_2) + 2x y_1^2 = n^2 (-2 y y_1)$.Dividing by $2 y_1$ (assuming $y_1 
eq 0$):$x^2 y_2 + x y_1 = -n^2 y \Rightarrow x^2 y_2 + x y_1 + n^2 y = 0$.

Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}$. Then

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