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 :

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 :

${2 \over {\sqrt 5 }}$

$ - {{\sqrt 5 } \over 2}$

${{\sqrt 5 } \over 2}$

$ - {{\sqrt 5 } \over 4}$

Solution

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