Let y = y(x) be a solution of the differential equation, $\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$, |x| < 1. If $y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$, then $y\left( { - {1 \over {\sqrt 2 }}} \right)$ is equal to :

Question

Let y = y(x) be a solution of the differential equation,

$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$, |x| < 1.

If $y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$, then $y\left( { - {1 \over {\sqrt 2 }}} \right)$ is equal to :

$ - {{\sqrt 3 } \over 2}$

None of those

${{1 \over {\sqrt 2 }}}$

$-{{1 \over {\sqrt 2 }}}$

Solution

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