Question

If \(\sin ^{-1}\left(\frac{y}{2}\right)=\tan ^{-1}\left(\frac{x}{2}\right),\ |y|\lt 2\), such that \(\frac{dx}{dy}=k(4+x^{2})\), then \(k=\ldots\)

• A. \(\frac{x}{2y}\)
• B. \(\frac{x}{4y}\)
• C. \(\frac{x}{y}\)
• D. \(\frac{y}{x}\)

Hint:

Put both sides equal to one angle, express x and y in terms of that angle and differentiate x with respect to y.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(\sin ^{-1}\left(\frac{y}{2}\right)=\tan ^{-1}\left(\frac{x}{2}\right)=\theta\).
Then \(y=2\sin \theta\) and \(x=2\tan \theta\).

Step 2: Differentiate
\(\frac{dx}{d\theta }=2\sec ^{2}\theta\) and \(\frac{dy}{d\theta }=2\cos \theta\).
So \(\frac{dx}{dy}=\frac{\sec ^{2}\theta }{\cos \theta }=\sec ^{3}\theta\).

Step 3: Use 4 + x^2
\(4+x^{2}=4+4\tan ^{2}\theta =4\sec ^{2}\theta\).
Hence \(k=\frac{\sec ^{3}\theta }{4\sec ^{2}\theta }=\frac{\sec \theta }{4}\).

Step 4: Express in x and y
\(\frac{x}{y}=\frac{2\tan \theta }{2\sin \theta }=\sec \theta\), so \(k=\frac{x}{4y}\).

Final Answer:
The constant is \(k=\frac{x}{4y}\). \[ \boxed{\text{(B) }\dfrac{x}{4y}} \]