Let
$$f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {{{{n^n}\left( {x + n} \right)\left( {x + {n \over 2}} \right)...\left( {x + {n \over n}} \right)} \over {n!\left( {{x^2} + {n^2}} \right)\left( {{x^2} + {{{n^2}} \over 4}} \right)....\left( {{x^2} + {{{n^2}} \over {{n^2}}}} \right)}}} \right)^{{x \over n}}},$$ for

all $$x>0.$$ Then

Let f: R $$ \to \left( {0,\infty } \right)$$ and g : R $$ \to $$ R be twice differentiable functions such that f'' and g'' are continuous functions on R. Suppose f'$$(2)$$ $$=$$ g$$(2)=0$$, f''$$(2)$$$$ \ne 0$$ and g'$$(2)$$ $$ \ne 0$$. If
$$\mathop {\lim }\limits_{x \to 2} {{f\left( x \right)g\left( x \right)} \over {f'\left( x \right)g'\left( x \right)}} = 1,$$ then

Let $$P$$ be the point on the parabola $${y^2} = 4x$$ which is at the shortest distance from the center $$S$$ of the circle $${x^2} + {y^2} - 4x - 16y + 64 = 0$$. Let $$Q$$ be the point on the circle dividing the line segment $$SP$$ internally. Then

Let $${F_1}\left( {{x_1},0} \right)$$ and $${F_2}\left( {{x_2},0} \right)$$ for $${{x_1} < 0}$$ and $${{x_2} > 0}$$, be the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. Suppose a parabola having vertex at the origin and focus at $${F_2}$$ intersects the ellipse at point $$M$$ in the first quadrant and at point $$N$$ in the fourth quadrant.

The orthocentre of the triangle $${F_1}MN$$ is