Let tangent to $${y^2} = 4x$$ be $$y = mx + {1 \over m}$$

Since this is also tangent to $${x^2} = - 32y$$

$$\therefore$$ $${x^2} = - 32\left( {mx + {1 \over m}} \right)$$

$$ \Rightarrow {x^2} + 32mx + {{32} \over m} = 0$$

Now, $$D=0$$

$${\left( {32} \right)^2} - 4\left( {{{32} \over m}} \right) = 0$$

$$ \Rightarrow {m^3} = {4 \over {32}} \Rightarrow m = {1 \over 2}$$

Given $$e{q^n}$$ of ellipse can be written as

$${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$$

Now, equation of any variable tangent is

$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$$

where $$m$$ is slope of the tangent

So, equation of perpendicular line drawn-

from center to tangent is

$$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Eliminating $$m,$$ we get

$$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$$

$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$$

$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$

Let common tangent be

$$y = mx + {{\sqrt 5 } \over m}$$

Since, perpendicular distance from center of the circle to

the common tangent is equal to radius of the circle,

therefore $${{{{\sqrt 5 } \over m}} \over {\sqrt {1 + {m^2}} }} = \sqrt {{5 \over 2}} $$

On squaring both the side, we get

$${m^2}\left( {1 + {m^2}} \right) = 2$$

$$ \Rightarrow {m^4} + {m^2} - 2 = 0$$

$$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$

$$ \Rightarrow m = \pm 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ ( as $$m \ne \pm \sqrt 2 $$ )

$$y = \pm \left( {x + \sqrt 5 } \right),$$ both statements are correct as $$m = \pm 1$$

satisfies the given equation of statement - $$2.$$

From the given equation of ellipse, we have

$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$

$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$

Now, radius of this circle $$ = {a^2} = 16$$

$$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$$

Now equation of circle is

$${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$

$${x^2} + y{}^2 - 6y - 7 = 0$$