$$e = \frac{1}{2}.$$   Directrix, $$x = \frac{a}{e} = 4$$
$$\therefore a = 4 \times \frac{1}{2} = 2\,\,\,\,\,\,\,\,\,\,\,\therefore b = 2\sqrt {1 - \frac{1}{4}} = \sqrt 3 $$
Equation of ellipse is $$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1\,\,\,\, \Rightarrow 3{x^2} + 4{y^2} = 12$$

$$\eqalign{ & \frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}} \cr & a = \sqrt {\frac{{144}}{{25}}} ,\,b = \sqrt {\frac{{81}}{{25}}} ,\,e = \sqrt {1 + \frac{{81}}{{144}}} = \frac{{15}}{{12}} = \frac{5}{4} \cr} $$
$$\therefore $$ Foci $$ = \left( { \pm 3,\,0} \right)$$
$$\therefore $$ foci of ellipse $$=$$ foci of hyperbola
$$\therefore $$ for ellipse $$ae = 3$$   but $$a=4,$$
$$\therefore e = \frac{3}{4}$$
Then $${b^2} = {a^2}{\left( {1 - e} \right)^2}\,\,\,\,\,\,\, \Rightarrow {b^2} = 16\left( {1 - \frac{9}{{16}}} \right) = 7$$

The tangent at $$\left( {4\cos \,\phi + 2\sin \,\phi } \right)$$    is $$4\cos \,\phi .x + 4.2\sin \,\phi .y = 16$$
Being normal, it passes through the centre $$\left( {4,\,2} \right)$$
So, $$16\cos \,\phi + 16\sin \,\phi = 16$$     or $$\cos \,\phi + \sin \,\phi = 1$$
It is satisfied by $$\phi = \frac{\pi }{2}$$

As rectangle $$ABCD$$   circumscribed the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$$
$$\therefore A = \left( {3,\,2} \right)$$

Let the ellipse circumscribing the rectangle $$ABCD$$   is $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$
Given that it passes through $$\left( {a,\,4} \right)$$
$$\therefore {b^2} = 16$$
Also it passes through $$A\left( {3,\,2} \right)$$
$$\eqalign{ & \therefore \frac{9}{{{a^2}}} + \frac{4}{{16}} = 1\,\,\, \Rightarrow {a^2} = 12 \cr & \therefore e = \sqrt {1 - \frac{{12}}{{16}}} = \sqrt {\frac{1}{4}} = \frac{1}{2} \cr} $$

The point is $$\left( {{x_1},\,{y_1}} \right)$$  if $$x{x_1} + 3y{y_1} = 9$$    has the slope $$1,$$ i.e., $$ - \frac{{{x_1}}}{{3{y_1}}} = 1$$   and $$x_1^2 + 3y_1^2 = 9.$$   Solve the two equations.

Equation of circle is $${\left( {x - 1} \right)^2} + {y^2} = 1$$
$$ \Rightarrow $$ radius $$=1$$  and diameter $$=2$$
$$\therefore $$ Length of semi-minor axis is $$2$$
Equation of circle is $${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$$
$$ \Rightarrow $$ radius $$=2$$  and diameter $$=4$$
$$\therefore $$ Length of semi major axis is $$4$$
We know, equation of ellipse is given by
$$\eqalign{ & \frac{{{x^2}}}{{{{\left( {Major\,axis} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {Minor\,axis} \right)}^2}}} = 1 \cr & \Rightarrow \frac{{{x^2}}}{{{{\left( 4 \right)}^2}}} + \frac{{{y^2}}}{{{{\left( 2 \right)}^2}}} = 1 \cr & \Rightarrow \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{4} = 1 \cr & \Rightarrow {x^2} + 4{y^2} = 16 \cr} $$

Here, the situation is standard. The distance between foci $$ = 2ae = 4$$   and $$e = \frac{2}{3}$$
$$ \Rightarrow \,\,a = 3$$   So, $${b^2} = {a^2}\left( {1 - {e^2}} \right) = 9.\left( {1 - \frac{4}{9}} \right) = 5$$

The tangent is $$\frac{{3\sqrt 3 \cos \,\phi .x}}{{27}} + \frac{{\sin \,\phi .y}}{1} = 1{\text{ or }}\frac{x}{{3\sqrt 3 \sec \,\theta }} + \frac{y}{{{\text{cosec}}\,\theta }} = 1$$
$$\therefore \,\,z = $$  sum of intercepts on axes $$ = 3\sqrt 3 \sec \,\theta + {\text{cosec}}\,\theta $$
For maximum or minimum, $$\frac{{dz}}{{d\theta }} = 0\,\,\, \Rightarrow {\cot ^3}\theta = 3\sqrt 3 \,\,\, \Rightarrow \cot \,\theta = \sqrt 3 \,\,\, \Rightarrow \theta = \frac{\pi }{6}$$

$$\eqalign{ & \frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1 \cr & {\text{Put }}x = 3 \cr} $$

$$\eqalign{ & \frac{9}{{25}} + \frac{{{y^2}}}{9} = 1 \cr & y = \frac{{12}}{5} \cr & P = \left( {3,\,\frac{{12}}{5}} \right) \cr & r = PO = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - \frac{{12}}{5}} \right)}^2}} = \frac{{17}}{5} \cr & s = PO' = \sqrt {\left[ { - 4 - 3} \right] + {{\left( {0 - \frac{{12}}{5}} \right)}^2}} = \frac{{33}}{5} \cr & r + s = \frac{{17}}{5} + \frac{{33}}{5} = \frac{{50}}{5} = 10{\text{ units}} \cr} $$

$$\sqrt {{a^2} - {b^2}} = \pm ae\left( {0,\,ae} \right)$$
So, the points are $$\left( {ae,\,0} \right)$$  and $$\left( {0,\, - ae} \right)$$
Let $$\frac{x}{a}\cos \,\theta + \frac{y}{b}\sin \,\theta = 1$$     be a tangent then sum of squares of perpendicular from these points is
$$\eqalign{ & = \frac{{{{\left( {1 - \frac{{ae}}{b}\sin \,\theta } \right)}^2} + {{\left( {1 + \frac{{ae}}{b}\sin \,\theta } \right)}^2}}}{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}} \cr & = \frac{{2\left( {1 + \frac{{{a^2}{e^2}}}{{{b^2}}}{{\sin }^2}\theta } \right)}}{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}} \cr & = 2{a^2}\left( {\frac{{{b^2} + \left( {{a^2} - {b^2}} \right){{\sin }^2}\theta }}{{{b^2}\left( {1 - \sin \,\theta } \right) + {a^2}{{\sin }^2}\theta }}} \right) \cr & = 2{a^2} \cr} $$