Let the coordinates of $$Q$$ be $$\left( {h,\,k} \right).$$  Since the point $$R$$ lies on the parabola. let its coordinates be $$\left( {a{t^2},\,2at} \right).$$

Since $$R$$ is mid point of $$PQ,$$
$$\eqalign{ & \therefore \,a{t^2} = \frac{{{x_1} + h}}{2}{\text{ and }}2at = \frac{{{y_1} + k}}{2} \cr & \Rightarrow {t^2} = \frac{{{x_1} + h}}{{2a}}{\text{ and }}t = \frac{{{y_1} + k}}{{4a}} \cr} $$
Equating the two values of $$t,$$ we get
$${\left( {\frac{{{y_1} + k}}{{4a}}} \right)^2} = \frac{{{x_1} + h}}{{2a}}\, \Rightarrow {\left( {{y_1} + k} \right)^2} = 8a\left( {{x_1} + h} \right)$$
Hence, locus of $$Q\left( {h,\,k} \right)$$  is $${\left( {y + {y_1}} \right)^2} = 8a\left( {x + {x_1}} \right)$$

Equation of the normal to a parabola $${y^2} = 4bx$$   at point $$\left( {bt_1^2,\,2b{t_1}} \right)$$   is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$
As given, it also passes through $$\left( {bt_2^2,\,2b{t_2}} \right)$$   then
$$\eqalign{ & 2b{t_2} = - {t_1}bt_2^2 + 2b{t_1} + bt_1^3 \cr & \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right) \cr & \Rightarrow {t_2} + {t_1} = - \frac{2}{{{t_1}}} \cr & \Rightarrow {t_2} = - {t_1} - \frac{2}{{{t_1}}} \cr} $$

Let the parabola be $${y^2} = 4ax$$   and $$P = \left( {at_1^2,\,2a{t_1}} \right),\,P' = \left( {at_2^2,\,2a{t_2}} \right)$$
$$\therefore \,VM = at_1^2,\,VM' = at_2^2{\text{ and }}VO = k,{\text{ where}}$$
\[\left| \begin{array}{l} \,a_1^2\,\,\,\,\,\,2a{t_1}\,\,\,\,\,\,1\\ at_2^2\,\,\,\,\,2a{t_2}\,\,\,\,\,1\\ \,\,\,k\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\]
$$\eqalign{ & \left( {\because P,\,P',\,O\,{\text{are collinear}}} \right)\,\,\, \Rightarrow k + a{t_1}{t_2} = 0 \cr & \therefore \,VM.VM' = {\left( {a{t_1}{t_2}} \right)^2} = {k^2} = O{V^2} \cr} $$

The graph shows $$\lambda > 0.$$

$$\left( {a,\,2a} \right)$$  is an interior point of $${y^2} - 16x = 0$$    if $${\left( {2a} \right)^2} - 16a < 0,$$    i.e., $${a^2} - 4a < 0$$
$$V\left( {0,\,0} \right)$$  and $$\left( {a,\,2a} \right)$$  are on the same side of $$x - 4 = 0.$$  So, $$a - 4 < 0,$$   i.e., $$a < 4$$
Now, $${a^2} - 4a < 0\,\,\,\,\,\, \Rightarrow 0 < a < 4$$

The locus is the directrix, which is a line.

The equation of the chord is $$y - 2{t_1} = \frac{2}{{{t_1} + {t_2}}}\left( {x - t_1^2} \right)$$
The equation of the tangent at $$'{t_1}'$$ is $$y.2{t_1} = 2\left( {x + t_1^2} \right).$$    Therefore, the slope of the normal at $$'{t_1}'$$ is $$ - {t_1}.$$   So, the chord will be the normal at $$'{t_1}'$$ if $$\frac{2}{{{t_1} + {t_2}}} = - {t_1}$$

Let the ends of the focal chord be $$\left( {at_1^2,\,2a{t_1}} \right)$$   and $$\left( {at_2^2,\,2a{t_2}} \right)$$
So, $${t_1}{t_2} + 1 = 0.$$
Then $${c^2} = {\left\{ {a\left( {t_1^2 - t_2^2} \right)} \right\}^2} + {\left\{ {2a\left( {{t_1} - {t_2}} \right)} \right\}^2}$$
The equation of the focal chord is
$$\eqalign{ & y - 2a{t_1} = \frac{{2a{t_1} - 2a{t_2}}}{{at_1^2 - at_2^2}}\left( {x - at_1^2} \right) \cr & {\text{or }}y - 2a{t_1} = \frac{2}{{{t_1} + {t_2}}}\left( {x - at_1^2} \right) \cr & {\text{So, }}b = \frac{{\frac{2}{{{t_1} + {t_2}}}\left( { - at_1^2} \right) + 2a{t_1}}}{{\sqrt {1 + \frac{4}{{{{\left( {{t_1} + {t_2}} \right)}^2}}}} }} = \frac{{2a{t_1}{t_2}}}{{\sqrt {4 + {{\left( {{t_1} + {t_2}} \right)}^2}} }} \cr & \therefore \,b = \frac{{ - 2a}}{{\sqrt {2 + t_1^2 + t_2^2} }} \cr & {\text{Also, }}{c^2} = {a^2}{\left( {{t_1} - {t_2}} \right)^2}\left\{ {{{\left( {{t_1} + {t_2}} \right)}^2} + 4} \right\} = {a^2}\left( {t_1^2 + t_2^2 + 2} \right)\left( {t_1^2 + t_2^2 + 2} \right) \cr & \therefore \,\,c = a\left( {2 + t_1^2 + t_2^2} \right) \cr & {\text{Hence, }}{b^2} = \frac{{4{a^2}}}{{2 + t_1^2 + t_2^2}} = \frac{{4{a^2}}}{{\frac{c}{a}}} = \frac{{4{a^3}}}{c} \cr} $$

$$\eqalign{ & {\text{Solving }}x - 2y = 1,\,{y^2} = kx,\,{\text{ we get}}\, \cr & {y^2} = k\left( {1 + 2y} \right){\text{ or }}{y^2} - 2ky - k = 0 \cr & \therefore \,{y_1} + {y_2} = 2k,\,\,{y_1}.{y_2} = - k \cr & \therefore 16 = {\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{{y_1^2}}{k} - \frac{{y_2^2}}{k}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{y_1} - {y_2}} \right)^2}.\left\{ {\frac{{{{\left( {{y_1} + {y_2}} \right)}^2}}}{{{k^2}}} + 1} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {{{\left( {{y_1} + {y_2}} \right)}^2} - 4{y_1}.{y_2}} \right\}\left\{ {\frac{{4{k^2}}}{{{k^2}}} + 1} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5\left\{ {4{k^2} + 4k} \right\} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20{k^2} + 20k\,;\,\,\,\,\,\,\,\,\,\,\,\therefore \,5{k^2} + 5k - 4 = 0 \cr} $$

The tangents to the parabola $${y^2} = 4ax$$   at the points $$\left( {at_1^2,\,2a{t_1}} \right)$$   and $$\left( {at_2^2,\,2a{t_2}} \right)$$   meet at $$\left( {a{t_1}{t_2},\,a\left( {{t_1} + {t_2}} \right)} \right).$$    Here $$a = 1,\,{t_1} = 1,\,{t_2} = 2.$$     So they meet at $$\left( {2,\,3} \right),$$  which is on the line $$y = 3.$$