Question

For \(a \gt 0,\ t \in \left(0, \dfrac{\pi}{2}\right)\), let \(x = \sqrt{a^{\sin^{-1}t}}\) and \(y = \sqrt{a^{\cos^{-1}t}}\), Then, \(1 + \left(\dfrac{dy}{dx}\right)^2\) equals :

• A. \(\dfrac{x^2}{y^2}\)
• B. \(\dfrac{y^2}{x^2}\)
• C. \(\dfrac{x^2 + y^2}{y^2}\)
• D. \(\dfrac{x^2 + y^2}{x^2}\)

Hint:

Differentiate \(x\) and \(y\) with respect to \(t\); note \(\frac{d}{dt}\sin^{-1}t = -\frac{d}{dt}\cos^{-1}t\).

Answer 1

The Correct Option is D

\(x = a^{\frac12\sin^{-1}t}\) and \(y = a^{\frac12\cos^{-1}t}\).
\(\dfrac{dx}{dt} = x\cdot\dfrac{\ln a}{2}\cdot\dfrac{1}{\sqrt{1-t^2}}\) and \(\dfrac{dy}{dt} = y\cdot\dfrac{\ln a}{2}\cdot\left(-\dfrac{1}{\sqrt{1-t^2}}\right)\).
\(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = -\dfrac{y}{x}\).
\(1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{y^2}{x^2} = \dfrac{x^2 + y^2}{x^2}\).
Answer: (4) \(\dfrac{x^2 + y^2}{x^2}\)