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 :

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Answer: B. - {{\sqrt 5 } \over 2} y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} ....(1) On differentiating both side of eq. (1) w.r.t. x we get, {{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }} = 0 - \sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}} Put x = {1 \over 2} and y = - {1 \over 4}, we get {{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} ight){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}} = - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}} 	herefore {{dy} \over {dx}} = - {{\sqrt 5 } \over 2}

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 :

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