Here, $$a = \cos \,\alpha ,\,\,b = \sin \,\alpha $$
$$\eqalign{ & {b^2} = {a^2}\left( {{e^2} - 1} \right) \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right) \cr & \Rightarrow {e^2} - 1 = {\tan ^2}\alpha \cr & \Rightarrow e = \sec \,\alpha \cr} $$
So, $$ae = 1$$
$$\therefore $$  abscissae of foci $$ = \pm ae = \pm 1$$

For the ellipse, $${a^2} = 16;\,\,{b^2} = {a^2}\left( {1 - {e^2}} \right)$$
$$ \Rightarrow e = \frac{{\sqrt {16 - {b^2}} }}{4}\,\,\,\, \Rightarrow ae = \sqrt {{{16}^2} - {b^2}} $$
For the hyperbola, $${a^2} = \frac{{144}}{{25}},\,\,{b^2} = \frac{{81}}{{25}};\,\,{b^2} = {a^2}\left( {{e^2} - 1} \right) \Rightarrow e = \frac{5}{4} \Rightarrow ae = 3$$
$$\therefore \,\,\,\sqrt {16 - {b^2}} = 3$$

Equation of normal at $$P\left( {x,\,y} \right)$$   is $$Y - y = - \frac{{dx}}{{dy}}\left( {X - x} \right)$$
Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,\,0} \right)$$  (let)
$$\eqalign{ & \therefore 0 - y = - \frac{{dx}}{{dy}}\left( {X - x} \right) \cr & \Rightarrow y\frac{{dy}}{{dx}} = X - x \cr & \Rightarrow X = x + y\frac{{dy}}{{dx}} \cr} $$
$$\therefore $$ Co-ordinate of $$G\,\,\left( {x + y\frac{{dy}}{{dx}},\,0} \right)$$
Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$
$$\because $$ Distance cannot be $$-ve,$$  therefore abscissa $$x$$ should be $$+ve$$
$$\eqalign{ & \therefore x + y\frac{{dy}}{{dx}} = 2x \cr & \Rightarrow y\frac{{dy}}{{dx}} = x \cr & \Rightarrow ydy = xdx \cr} $$
On Integrating $$ \Rightarrow \frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + {c_1}\,\, \Rightarrow {x^2} - {y^2} = - 2{c_1}$$
$$\therefore $$ The curve is a hyperbola.

Equation of tangent at point $$'\varphi '$$ on the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$    is
$$\eqalign{ & \frac{x}{a}\sec \,\varphi - \frac{y}{b}\tan \,\varphi = 1 \cr & {\text{or }}y = \frac{b}{a}x\,{\text{cosec}}\,\varphi - b\,\cot \,\varphi ......\left( 1 \right) \cr & {\text{If }}y = mx + \sqrt {{a^2}{m^2} - {b^2}} ......\left( 2 \right) \cr} $$
also touches the hyperbola then on comparing $$\left( 1 \right)$$ & $$\left( 2 \right)$$
$$\eqalign{ & 1 = \frac{{\frac{b}{a}{\text{cosec}}\,\varphi }}{m} = \frac{{ - b\,\cot \,\varphi }}{{\sqrt {{a^2}{m^2} - {b^2}} }} \cr & {\text{Hence, }}m = \frac{b}{a}{\text{cosec}}\,\varphi \,; \cr & {\text{or cosec}}\,\varphi = \frac{{am}}{b} \cr & {\text{or }}\sin \,\varphi = \frac{b}{{am}} \cr & {\text{or }}\varphi = {\sin ^{ - 1}}\frac{b}{{am}} \cr} $$

$$\frac{4}{{{a^2}}} - \frac{9}{{{b^2}}} = 1$$   and $${b^2} = {a^2}\left( {{e^2} - 1} \right) = {a^2}.3$$
So, $$\frac{4}{{{a^2}}} - \frac{9}{{3{a^2}}} = 1\,\,\,\,\, \Rightarrow {a^2} = 1$$
Transverse axis $$ = 2a = 2$$

Here equation of hyperbola is $$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{36}} = 1$$
Now, $$PQ$$  is the chord of content
$$\therefore $$ Equation of $$PQ$$  is :
$$\eqalign{ & \frac{{x\left( 0 \right)}}{9} - \frac{{y\left( 3 \right)}}{{36}} = 1 \cr & \Rightarrow y = - 12 \cr} $$

$$\eqalign{ & \therefore {\text{Area of }}\Delta PQT = \frac{1}{2} \times TR \times PQ \cr & \because P \equiv \left( {3\sqrt 5 ,\, - 12} \right) \cr & \therefore TR = 3 + 12 = 15 \cr & \therefore {\text{Area of }}\Delta PQT = \frac{1}{2} \times 15 \times 6\sqrt 5 = 45\sqrt 5 \,\,{\text{sq}}{\text{. units}} \cr} $$

Since, $$lx + my + n = 0$$    is a normal to $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,$$   then $$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{\left( {{a^2} + {b^2}} \right)}^2}}}{{{n^2}}}$$     but it is given that $$mx - y + 7\sqrt 3 $$    is normal to hyperbola $$\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1$$   then $$\frac{{24}}{{{m^2}}} - \frac{{18}}{{{{\left( { - 1} \right)}^2}}} = \frac{{{{\left( {24 + 18} \right)}^2}}}{{{{\left( {7\sqrt 3 } \right)}^2}}}\,\, \Rightarrow m = \frac{2}{{\sqrt 5 }}$$

KEY CONCEPT :
Equation of the normal to the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$    at the point $$\left( {a\,\sec \,\alpha ,\,b\,\tan \,\alpha } \right)$$    is given by $$ax\,\cos \,\alpha + by\,\cot \,\alpha = {a^2} + {b^2}$$
Normal at $$\theta ,\,\phi $$  are \[\left\{ \begin{array}{l} ax\,\cos \,\theta + by\,\cot \,\theta = {a^2} + {b^2}\\ ax\,\cos \,\theta + by\,\cot \,\phi = {a^2} + {b^2} \end{array} \right.\]
where $$\phi = \frac{\pi }{2} - \theta $$   and these pass through $$\left( {h,\,k} \right)$$
$$\eqalign{ & \therefore ah\,\cos \,\theta + bk\,\cot \,\theta = {a^2} + {b^2} \cr & \,\,\,\,ah\,\sin \,\theta + bk\,\tan \,\theta = {a^2} + {b^2} \cr} $$
Eliminating $$h,\,bk\left( {\cot \,\theta \,\sin \,\theta - \tan \,\theta \cos \,\theta } \right)$$
$$ = \left( {{a^2} + {b^2}} \right)\left( {\sin \,\theta - \cos \,\theta } \right)\,\,\,\,{\text{or }}\,k = - \frac{{\left( {{a^2} + {b^2}} \right)}}{b}$$

The given hyperbola is
$$\eqalign{ & {x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0 \cr & \Rightarrow \left( {{x^2} - 2\sqrt 2 x + 2} \right) - 2\left( {{y^2} + 2\sqrt 2 y + 2} \right) = 6 + 2 - 4 \cr & \Rightarrow {\left( {x - \sqrt 2 } \right)^2} - 2{\left( {y + \sqrt 2 } \right)^2} = 4 \cr & \Rightarrow \frac{{{{\left( {x - \sqrt 2 } \right)}^2}}}{{{2^2}}} - \frac{{{{\left( {y + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}} = 1 \cr & \therefore a = 2,\,\,\,b = \sqrt 2 \cr & \Rightarrow e = \sqrt {1 + \frac{2}{4}} = \sqrt {\frac{3}{2}} \cr} $$
Clearly $$\Delta ABC$$   is a right triangle.

$$\eqalign{ & \therefore Ar\left( {\Delta ABC} \right) = \frac{1}{2} \times AC \times BC = \frac{1}{2}\left( {ae - a} \right) \times \frac{{{b^2}}}{a} \cr & = \frac{1}{2}\left( {e - 1} \right) \times {b^2} = \frac{1}{2}\left( {\sqrt {\frac{3}{2}} - 1} \right) \times 2 = \sqrt {\frac{3}{2}} - 1 \cr} $$

The equation of tangent at a point $$\left( {{x_1},\,{y_1}} \right)$$   on the hyperbola $${x^2} - {y^2} = c$$   is given by $$x{x_1} - y{y_1} = c$$
Chord $$x = 9$$  meets $${x^2} - {y^2} = 9$$   at $$\left( {9,\,6\sqrt 2 } \right)$$   and $$\left( {9,\, - 6\sqrt 2 } \right)$$   at which tangents are
$$\eqalign{ & 9x - 6\sqrt 2 y = 9{\text{ and }}9x + 6\sqrt 2 y = 9 \cr & {\text{or, }}3x - 2\sqrt 2 y - 3 = 0{\text{ and }}3x + 2\sqrt 2 y - 3 = 0 \cr} $$
$$\therefore $$  Combined equation of tangents is
$$\eqalign{ & \left( {3x - 2\sqrt 2 y - 3} \right)\left( {3x + 2\sqrt 2 y - 3} \right) = 0 \cr & {\text{or, }}9{x^2} - 8{y^2} - 18x + 9 = 0 \cr} $$