Question

If \(sec^{-1}(\frac{x^2+y^2}{x^2-y^2}) = 2a\) such that \(y\frac{dy}{dx} = x\cdot f(a)\) then the value of \(f(\frac{2π}{3})\) is

• A. \(-3\)
• B. \(\sqrt{3}\)
• C. \(3\)
• D. \(\frac{1}{2}\)

Hint:

Convert the secant relation to tan a = y/x and differentiate with a constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Here \(a\) acts as a constant parameter. The relation \(\sec^{-1}\frac{x^2+y^2}{x^2-y^2} = 2a\) defines \(y\) as a function of \(x\), and we need \(y\,\frac{dy}{dx}\) in terms of \(x\) and a function of \(a\).

Step 2: Simplify the relation:
Taking secant, \(\frac{x^2+y^2}{x^2-y^2} = \sec 2a\), so \(\frac{x^2-y^2}{x^2+y^2} = \cos 2a\).
By componendo and dividendo, \(\frac{y^2}{x^2} = \frac{1-\cos 2a}{1+\cos 2a} = \tan^2 a\).
So \(y = x\tan a\).

Step 3: Differentiate:
\(\frac{dy}{dx} = \tan a\). Therefore
\[ y\frac{dy}{dx} = x\tan a \cdot \tan a = x\tan^2 a \]
So \(f(a) = \tan^2 a\).

Step 4: Evaluate:
\(f\left(\frac{2\pi}{3}\right) = \tan^2\frac{2\pi}{3} = (-\sqrt{3})^2 = 3\).

Step 5: Why the other options are wrong.
Option (A) \(-3\) cannot occur because \(\tan^2\) is never negative. Option (B) \(\sqrt3\) is \(|\tan 60^{\circ}|\) without squaring. Option (D) \(1/2\) has no connection to this function.

Final Answer:
\(f(2\pi/3) = 3\), option (C). \[ \boxed{3} \]