The equation of the pair of lines $$VA$$  and $$VB,$$  where $$V$$ is the vertex $$ = \left( {0,\,0} \right)$$   is $${y^2} = 4ax.\frac{{y - mx}}{c}{\text{ or }}c{y^2} - 4axy + 4am{x^2} = 0$$
The pair is at right angles if $$c + 4am = 0.$$

Let the equation of tangent to $${y^2} = 4x$$   be $$y = mx + \frac{1}{m}$$   where $$m$$ is the slope of the tangent.
If it is tangent to the circle $${\left( {x - 3} \right)^2} + {y^2} = 9$$     then length of perpendicular to tangent from centre $$\left( {3,\,0} \right)$$  should be equal to the radius $$3.$$
$$\eqalign{ & \therefore \frac{{3m + \frac{1}{m}}}{{\sqrt {{m^2} + 1} }} = 3 \cr & \Rightarrow 9{m^2} + \frac{1}{{{m^2}}} + 6 = 9{m^2} + 9 \cr & \Rightarrow m = \pm \frac{1}{{\sqrt 3 }} \cr} $$
$$\therefore $$ Tangents are $$x - y\sqrt 3 + 3 = 0$$    and $$x + y\sqrt 3 + 3 = 0$$    out of which $$x - y\sqrt 3 + 3 = 0$$    meets the parabola at $$\left( {3,\,2\sqrt 3 } \right)$$   i.e., above $$x$$-axis.

$$\eqalign{ & {\text{Clearly,}}\,\,A = \left( {8p,\,8p} \right){\text{ and }}B = \left( {8p,\, - 8p} \right) \cr & {\text{Also, }}V = \left( {0,\,0} \right) \cr & 'm'{\text{ of }}AV = 1;\,\,\,\,'m'{\text{ of }}BV = - 1 \cr & {\text{Therefore, }}AV \bot BV \cr} $$

Solving the equations,
$$\eqalign{ & {x^2} + \frac{{3\left( {x + 1} \right)}}{2} + 2x = 0 \cr & {\text{or, }}{x^2} + \frac{{7x}}{2} + \frac{3}{2} = 0 \cr & {\text{or, }}2{x^2} + 7x + 3 = 0 \cr & {\text{or, }}\left( {2x + 1} \right)\left( {x + 3} \right) = 0 \cr & {\text{or, }}x = - \frac{1}{2},\, - 3 \cr} $$
But $$x = - 3$$   makes $$y$$ imaginary because $$2{y^2} = 3\left( {x + 1} \right)$$
So, $$x = - \frac{1}{2}\,\,\,\,\,\, \Rightarrow y = \pm \frac{{\sqrt 3 }}{2}$$
$$\therefore $$  the length of the chord $$=$$ the distance between $$\left( { - \frac{1}{2},\,\frac{{\sqrt 3 }}{2}\,} \right)$$   and $$\left( { - \frac{1}{2},\, - \frac{{\sqrt 3 }}{2}\,} \right).$$

$$\eqalign{ & {\text{Here, }}S \equiv {y^2} - 4x = 0 \cr & S\left( {1,\,3} \right) = {3^2} - 4.1 > 0\,\,\,\,\,\, \Rightarrow P\left( {1,\,3} \right){\text{ is an exterior point}}{\text{.}} \cr & S\left( {1,\,1} \right) = {1^2} - 4.1 < 0\,\,\,\,\,\, \Rightarrow Q\left( {1,\,1} \right){\text{ is an interior point}}{\text{.}} \cr} $$

Since, the equation of tangent to parabola $${y^2} = 4x$$   is
$$y = mx + \frac{1}{m}.....(1)$$
The line (a) is also the tangent to circle
$${x^2} + {y^2} - 6x = 0$$
Then centre of circle $$ = \left( {3,\,0} \right)$$
radius of circle $$= 3$$
The perpendicular distance from centre to tangent is equal to the radius of circle
$$\eqalign{ & \therefore \frac{{\left| {3m + \frac{1}{m}} \right|}}{{\sqrt {1 + {m^2}} }} = 3 \cr & \Rightarrow {\left( {3m + \frac{1}{m}} \right)^2} = 9\left( {1 + {m^2}} \right) \cr & \Rightarrow m = \pm \frac{1}{{\sqrt 3 }} \cr} $$
Then, from equation (1) : $$y = \pm \frac{1}{{\sqrt 3 }}x \pm \sqrt 3 $$
Hence, $$\sqrt 3 y = x + 3$$   is one of the required common tangent.

$$\because {y^2} = 16x\,\, \Rightarrow a = 4$$
One end of focal of the parabola is at (1, 4)
$$\because \,y$$ co-ordinate of focal chord is $$2\,at$$
$$\eqalign{ & \therefore 2\,at = 4 \cr & \Rightarrow t = \frac{1}{2} \cr} $$
Hence, the required length of focal chord
$$ = a{\left( {t + \frac{1}{t}} \right)^2} = 4 \times {\left( {2 + \frac{1}{2}} \right)^2} = 25$$

Since, distance of vertex from origin is $$\sqrt 2 $$ and focus is $$2\sqrt 2 $$
$$\therefore $$ Vertex is $$\left( {1,\,1} \right)$$  and focus is $$\left( {2,\,2} \right),$$  directrix $$x+y=0$$

$$\therefore $$ Equation of parabola is
$$\eqalign{ & {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\frac{{x + y}}{{\sqrt 2 }}} \right)^2} \cr & \Rightarrow 2\left( {{x^2} - 4x + 4} \right) + 2\left( {{y^2} - 4y + 4} \right) = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 8\left( {x + y - 2} \right) \cr & \Rightarrow {\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right) \cr} $$

Let the parabola be $${y^2} = 4ax.\,\,Q$$    is the intersection of the lines $$x = 0$$  and $$ty = x + a{t^2},$$    where $$P = \left( {a{t^2},\,2at} \right).$$    Solving these, $$Q = \left( {0,\,at} \right).$$   Also $$S = \left( {a,\,0} \right).$$
$$\eqalign{ & \therefore \,\,S{P^2} = {\left\{ {a\left( {{t^2} - 1} \right)} \right\}^2} + 4{a^2}{t^2} = {a^2}{\left( {{t^2} + 1} \right)^2} \cr & \,\,\,\,\,\,\,\,S{Q^2} = {a^2} + {a^2}{t^2} = {a^2}\left( {{t^2} + 1} \right){\text{ and }}SV = a \cr & \therefore \,S{Q^2} = SP.SV\, \cr} $$

The sum of the ordinates of the feet $$ = {y_1} + {y_2} + {y_3} = 0$$
$$\eqalign{ & \therefore \,2 + \left( { - 2} \right) + {y_3} = 0 \cr & \therefore \,{y_3} = 0 \cr} $$