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 :

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

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :

Let $$\overrightarrow a $$ , $$\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors such that
$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$. If $$\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $$ and
$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$, then the ordered pair, $$\left( {\lambda ,\overrightarrow d } \right)$$ is equal to :

The area (in sq. units) of the region
{(x, y) $$ \in $$ R² | 4x² $$ \le $$ y $$ \le $$ 8x + 12} is :