If the straight line $$y = mx + c$$ touches the parabola $${y^2} - 4ax + 4{a^3} = 0$$, then c is

A normal is drawn at the point P to the parabola $${y^2} = 8x$$, which is inclined at 60$$^\circ$$ with the straight line $$y = 8$$. Then the point P lies on the straight line

For each parabola y = x² + px + q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

The origin is shifted to (1, 2). The equation y² $$-$$ 8x $$-$$ 4y + 12 = 0 changes to y² = 4ax, then a is equal to