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