Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}$. Then
Options:
Solution:
1774 Upvotes
Verified Answer
The correct answer is:
$x^{2} y_{2}+x y_{1}+n^{2} y=0$
Hint:
$\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}=n \times \log \left(\frac{x}{n}\right)$
$\Rightarrow-\frac{1}{\sqrt{b^{2}-y^{2}}} \cdot y_{1}=\frac{n}{x}$
$\Rightarrow x^{2} y_{1}^{2}=n^{2}\left(b^{2}-y^{2}\right) \Rightarrow x^{2} y_{2}+x y_{1}+n^{2} y=0$
$\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}=n \times \log \left(\frac{x}{n}\right)$
$\Rightarrow-\frac{1}{\sqrt{b^{2}-y^{2}}} \cdot y_{1}=\frac{n}{x}$
$\Rightarrow x^{2} y_{1}^{2}=n^{2}\left(b^{2}-y^{2}\right) \Rightarrow x^{2} y_{2}+x y_{1}+n^{2} y=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.