The focus of parabola $${y^2} = 2px$$   is $$\left( {\frac{p}{2},\,0} \right)$$  and directrix $$x = - \frac{p}{2}$$

In the figure, we have supposed that $$\left. {p > 0} \right]$$
$$\therefore $$ Centre of circle is $$\left( {\frac{p}{2},\,0} \right)$$  and radius $$ = \frac{p}{2} + \frac{p}{2} = p$$
$$\therefore $$ Equation of circle is $${\left( {x - \frac{p}{2}} \right)^2} + {y^2} = {p^2}$$
For points of intersection of $${y^2} = 2px.....({\text{i}})$$
and $$4{x^2} + 4{y^2} - 4px - 3{p^2} = 0.....({\text{ii}})$$
can be obtained by solving (i) and (ii) as follows
$$\eqalign{ & 4{x^2} + 8px - 4px - 3{p^2} = 0 \cr & \Rightarrow \left( {2x + 3p} \right)\left( {2x - p} \right) = 0 \cr & \Rightarrow x = \frac{{ - 3p}}{2},\,\frac{p}{2} \cr & \Rightarrow y = - 3{p^2}\left( {{\text{not possible}}} \right),\,\,{p^2} \Rightarrow y = \pm p \cr} $$
$$\therefore $$ Required pts are $$\left( {\frac{p}{2},\,p} \right),\,\,\left( {\frac{p}{2},\, - p} \right)$$

The equation is $${\left( {x - 0} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {\frac{{x + 2}}{{\sqrt 1 }}} \right)^2}$$
$${\text{or }}{\left( {y - 1} \right)^2} = 4\left( {x + 1} \right)$$
Clearly, $$x = {t^2} - 1$$   and $$y = 2t + 1$$   satisfy it for all t.

KEY CONCEPT :
The equation $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$
represents a parabola if $$\Delta \ne 0$$  and $${h^2} = ab$$
where $$\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$$
Now we have $$x = {t^2} + t + 1$$    and $$y = {t^2} - t + 1$$
$$\frac{{x + y}}{2} = {t^2} + 1,\,\frac{{x - y}}{2} = t$$       ( Adding and subtracting values of $$x$$ and $$y$$ )
Eliminating $$t,\,\,\,2\left( {x + y} \right) = {\left( {x - y} \right)^2} + 4.....(1)$$
$$ \Rightarrow {x^2} - 2xy + {y^2} - 2x - 2y + 4 = 0.....(2)$$
Here, $$a = 1,\,h = - 1,\,b = 1,\,g = - 1,\,f = - 1,\,c = 4$$
$$\therefore \Delta \ne 0.\,\,{\text{and }}{h^2} = ab$$
Hence the given curve represents a parabola.

The locus of perpendicular tangents is directrix
i.e., $$x = - 1$$

$$\eqalign{ & S{A^2} = {\left( {a - at_1^2} \right)^2} + {\left( {0 - 2a{t_1}} \right)^2} \cr & = {a^2}{\left( {1 - t_1^2} \right)^2} + 4{a^2}t_1^2 \cr & = {a^2}{\left( {1 - t_1^2} \right)^2} \cr & {\text{Similarly, }}S{B^2} = {a^2}{\left( {1 + t_2^2} \right)^2} \cr & {\text{But }}A,\,S,\,B{\text{ are collinear}} \cr & {\text{Hence,}}\,\frac{{2a{t_1} - 0}}{{at_1^2 - a}} = \frac{{2a{t_2} - 0}}{{at_2^2 - a}} \cr & {\text{or, }}\frac{{2{t_1}}}{{t_1^2 - 1}} = \frac{{2{t_1}}}{{t_2^2 - 1}} \cr & {\text{or, }}{t_1}t_2^2 - {t_1} = t_1^2{t_2} - {t_2} \cr & {\text{or, }}{t_1}{t_2}\left( {{t_2} - {t_1}} \right) + \left( {{t_2} - {t_1}} \right) = 0 \cr & {\text{But }}{t_1} \ne {t_2}.\,\,{\text{So, }}{t_1}{t_2} + 1 = 0 \cr & {\text{HM of }}SA,\,SB = \frac{{2SA.SB}}{{SA + SB}} = \frac{{2.a\left( {1 + t_1^2} \right).a\left( {1 + t_2^2} \right)}}{{a\left( {1 + t_1^2} \right) + a\left( {1 + t_2^2} \right)}} \cr & = \frac{{2{a^2}\left\{ {1 + \left( {t_1^2 + t_2^2} \right) + t_1^2t_2^2} \right\}}}{{a\left\{ {2 + \left( {t_1^2 + t_2^2} \right)} \right\}}} \cr & = \frac{{2{a^2}\left\{ {1 + \left( {t_1^2 + t_2^2} \right) + 1} \right\}}}{{a\left\{ {2 + \left( {t_1^2 + t_2^2} \right)} \right\}}} \cr & = 2a \cr} $$

Let $$y = Y,\,x - 1 = X.$$    Then the equation is $${Y^2} = 4X$$
So, the focus $$ = {\left( {1,\,0} \right)_{X,\,Y}} = \left( {2,\,0} \right)$$
Any line through the focus is $$y = m\left( {x - 2} \right).$$   Solving this with $${y^2} = 4\left( {x - 1} \right),\,{m^2}{\left( {x - 2} \right)^2} = 4\left( {x - 1} \right){\text{ or }}{m^2}{x^2} - 4\left( {{m^2} + 1} \right)x + 4\left( {{m^2} + 1} \right) = 0$$
If $$m \ne 0,\,D = 16{\left( {{m^2} + 1} \right)^2} - 16{m^2}\left( {{m^2} + 1} \right) = 16\left( {{m^2} + 1} \right) > 0{\text{ for all }}m$$
But, if $$m = 0$$  then $$x$$ does not have two real distinct values.
So $$m\, \in \,R$$  except $$m = 0$$

Given equation of ellipse is $$2{x^2} + {y^2} = 4$$
$$ \Rightarrow \frac{{2{x^2}}}{4} + \frac{{{y^2}}}{4} = 1\,\,\,\,\, \Rightarrow \frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1$$
Equation of tangent to the ellipse $$\frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1$$    is
$$y = mx \pm \sqrt {2{m^2} + 4} .....(1)$$
($$\because $$ equation of tangent to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$    is $$y=mx+c$$   where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}} $$    )
Now, Equation of tangent to the parabola
$${y^2} = 16\sqrt 3 x$$   is $$y = mx + \frac{{4\sqrt 3 }}{m}.....(2)$$
($$\because $$ equation of tangent to the parabola $${y^2} = 4ax$$   is $$y = mx + \frac{a}{m}$$   )
On comparing (1) and (2), we get
$$\frac{{4\sqrt 3 }}{m} = \pm \sqrt {2{m^2} + 4} $$
Squaring on both the sides, we get
$$\eqalign{ & 16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2} \cr & \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \cr & \Rightarrow 2{m^4} + 4{m^2} - 48 = 0 \cr & \Rightarrow {m^4} + 2{m^2} - 24 = 0 \cr & \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0 \cr & \Rightarrow {m^2} = 4\left( {\because {m^2} \ne - 6} \right) \cr & \Rightarrow m = \pm 2 \cr} $$
$$ \Rightarrow $$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3 $$
Thus, statement-1 is true.
Statement-2 is obviously true.

Consider two points $$P\left( {at_1^2,\,2a{t_1}} \right)$$   and $$Q\left( {at_2^2,\,2a{t_2}} \right)$$   on the parabola $${y^2} = 4ax$$
Given : $$\frac{{at_1^2}}{{at_2^2}} = \frac{{{m^2}}}{1}{\text{ or }}{t_1} = m{t_2}......\left( 1 \right)$$
Let $$R\left( {h,\,k} \right)$$  be the point of intersection of tangents at $$P$$ and $$Q.$$
Then, $$h = a{t_1}{t_2}$$   and $$k = a\left( {{t_1} + {t_2}} \right)$$
$$\eqalign{ & \Rightarrow h = amt_2^2{\text{ and }}k = a\left( {m{t_2} + {t_2}} \right)\,\,\,\,\,\,\,\,\left[ {{\text{using }}\left( 1 \right)} \right] \cr & \Rightarrow t_2^2 = \frac{h}{{am}}{\text{ and }}{t_2} = \frac{k}{{a\left( {m + 1} \right)}} \cr} $$
Equating the two values of $${t_2},$$ we get
$$\eqalign{ & \frac{{{k^2}}}{{{a^2}{{\left( {m + 1} \right)}^2}}} = \frac{h}{{am}} \cr & \Rightarrow {k^2} = ah\frac{{{{\left( {m + 1} \right)}^2}}}{m} \cr & \Rightarrow {k^2} = ah{\left( {\sqrt m + \frac{1}{{\sqrt m }}} \right)^2} \cr} $$
$$\therefore $$  Required locus is $${y^2} = ax{\left( {{m^{\frac{1}{2}}} + {m^{ - \frac{1}{2}}}} \right)^2}.$$

The given curve is $$y = {x^2} + 6$$
Equation of tangent at $$\left( {1,\,7} \right)$$  is
$$\eqalign{ & \frac{1}{2}\left( {y + 7} \right) = x.1 + 6 \cr & \Rightarrow 2x - y + 5 = 0.....(1) \cr} $$
As given this tangent (1) touches the circle $${x^2} + {y^2} + 16x + 12y + c = 0$$       at $$Q$$
Centre of circle $$ = \left( { - 8,\, - 6} \right).$$

Then equation of $$CQ$$  which is perpendicular to (1) and passes through $$\left( { - 8,\, - 6} \right)$$  is $$y + 6 = - \frac{1}{2}\left( {x + 8} \right)$$
$$ \Rightarrow x + 2y + 20 = 0.....(2)$$
Now $$Q$$ is points of intersection of (1) and (2)
$$\therefore $$ Solving equation (1) & (2) we get
$$x=-6,\,\,y=-7$$
$$\therefore $$ Required points is $$\left( { - 6,\, - 7} \right)$$

Since, vertex and focus of given parabola is $$\left( {2,\,0} \right)$$  and $$\left( {4,\,0} \right)$$  respectively

Then, equation of parabola is
$$\eqalign{ & {\left( {y - 0} \right)^2} = 4 \times 2\left( {x - 2} \right) \cr & \Rightarrow {y^2} = 8x - 16 \cr} $$
Hence, the point $$\left( {8,\,6} \right)$$  does not lie on given parabola.