Official solution
\[y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} \] ....(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
\[{{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}\]
= \[ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}\]
\[ \therefore \] \[{{dy} \over {dx}} = - {{\sqrt 5 } \over 2}\]