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

The distance of point of intersection of the tangents to the parabola x = 4y $$-$$ y² drawn at the points where it is meet by Y-axis, from its focus is