Question

Let y = y(x) be a function of x satisfying

\( y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} \) where k is a constant and

\( y\left( {{1 \over 2}} \right) = - {1 \over 4} \). Then \( {{dy} \over {dx}} \) at x = \( {1 \over 2} \), is equal to :

• A. \( {2 \over {\sqrt 5 }} \)
• B. \( - {{\sqrt 5 } \over 2} \)
• C. \( {{\sqrt 5 } \over 2} \)
• D. \( - {{\sqrt 5 } \over 4} \)

Hint:

Substitute \(x=\sin A,\ y=\sin B\): the relation becomes \(\sin(A+B)=k\), so \(A+B\) is locally constant.

Answer 1

The Correct Option is B

Step 1: Let \(x=\sin A,\;y=\sin B\). Then \(y\sqrt{1-x^2}+x\sqrt{1-y^2}=\sin B\cos A+\sin A\cos B=\sin(A+B)=k\).
Step 2: Hence \(\sin^{-1}x+\sin^{-1}y\) is constant (locally), so \(\dfrac{1}{\sqrt{1-x^2}}+\dfrac1{\sqrt{1-y^2}}\dfrac{dy}{dx}=0\), i.e. \(\dfrac{dy}{dx}=-\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\).
Step 3: At \(x=\tfrac12,\;y=-\tfrac14\): \(\sqrt{1-y^2}=\dfrac{\sqrt{15}}4,\;\sqrt{1-x^2}=\dfrac{\sqrt3}2\).
Step 4: \(\dfrac{dy}{dx}=-\dfrac{\sqrt{15}/4}{\sqrt3/2}=-\dfrac{\sqrt5}2\).
Answer: (B) \(-\tfrac{\sqrt5}{2}\)