The slope of the line touching both the parabolas $${y^2} = 4x$$ and $${x^2} = - 32y$$ is

Given : A circle, $$2{x^2} + 2{y^2} = 5$$ and a parabola, $${y^2} = 4\sqrt 5 x$$.
Statement-1 : An equation of a common tangent to these curves is $$y = x + \sqrt 5 $$.

Statement-2 : If the line, $$y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)$$ is their common tangent, then $$m$$ satiesfies $${m^4} - 3{m^2} + 2 = 0$$.

If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is

A parabola has the origin as its focus and the line $$x=2$$ as the directrix. Then the vertex of the parabola is at :