Tangent to y2 = 4x at (t2, 2t) is
y(2t) = 2(x + t2)
$$\Rightarrow$$ yt = x + t2 ...... (i)
Equation of normal at P(t2, 2t) is
y + tx = 2t + t3
Since, normal at P passes through centre of circle S(2, 8).
$$\therefore$$ 8 + 2t = 2t + t3 $$\Rightarrow$$ t = 2, i.e., P(4, 4)
$$\therefore$$ $$SP = \sqrt {{{(4 - 2)}^2} + {{(4 - 8)}^2}} = 2\sqrt 5 $$
$$\therefore$$ Option (a) is correct.
Also, SQ = 2
$$\therefore$$ PQ = SP $$-$$ SQ = 2$$\sqrt5$$ $$-$$ 2
Thus, $${{SQ} \over {QP}} = {1 \over {\sqrt 5 - 1}} = {{\sqrt 5 + 1} \over 4}$$
$$\therefore$$ Option (b) is incorrect.
Now, x-intercept of normal is
x = 2 + 22 = 6
$$\therefore$$ Option (c) is correct.
Slope of tangent $$ = {1 \over t} = {1 \over 2}$$
$$\therefore$$ Option (d) is correct.
In case of option (A)
x2 + y2 = 3 and x2 = 2y
Solving we get $$P(\sqrt 2 ,1)$$
Equation of tangent at P on the circle
$$x\sqrt 2 + y = 3$$ $$\therefore$$ $$\tan \alpha = - \sqrt 2 $$
Again, $${{{Q_2}{R_2}} \over {{Q_2}M}} = \sin \left( {\alpha - {\pi \over 2}} \right)$$
or, $${{2\sqrt 3 } \over {{Q_2}M}} = - \cos \alpha = - {1 \over {\sqrt 3 }}$$ [$$\because$$ $$\tan \alpha = - \sqrt 2 $$]
or, $${Q_2}M = 6$$
Similarly, $${Q_3}M = 6$$
$$\therefore$$ $${Q_2}{Q_3} = 12$$
In case of option (B)
$${{M{R_2}} \over {{Q_2}{R_2}}} = \tan \left( {\alpha - {\pi \over 2}} \right)$$
or, $${{M{R_2}} \over {2\sqrt 3 }} = - \tan \alpha = \sqrt 2 $$
or, $$M{R_2} = 2\sqrt 6 $$
Similarly, $$M{R_3} = 2\sqrt 6 $$ $$\therefore$$ $${R_2}{R_3} = 4\sqrt 6 $$
In case of option (C)
OP is the perpendicular drawn from O to R2R3
$$\therefore$$ area of $$\Delta O{R_2}{R_3} = {1 \over 2} \times OP \times {R_2}{R_3} = {1 \over 2} \times \sqrt 3 \times 4\sqrt 6 =6\sqrt 2 $$
In case of option (D)
PT is perpendicular drawn from P to Q2Q3
$$\therefore$$ area of $$\Delta P{Q_2}{Q_3} = {1 \over 2} \times PT \times {Q_2}{Q_3} = {1 \over 2} \times \sqrt 2 \times 12 = 6\sqrt 2 $$
Therefore, (A), (B), (C) are correct options.
Equation of family of circles touching hyperbola at (x1, y1) is
(x $$-$$ x1)2 + (y $$-$$ y1)2 + $$\lambda$$(xx1 $$-$$ yy1 $$-$$ 1) =0
Now, its centre is (x2, 0).
$$\therefore$$ $$\left[ {{{ - (\lambda {x_1} - 2{x_1})} \over 2},{{ - ( - 2{y_1} - \lambda {y_1})} \over 2}} \right] = ({x_2},0)$$
$$ \Rightarrow 2{y_1} + \lambda {y_1} = 0 \Rightarrow \lambda = - 2$$
and $$2{x_1} - \lambda x = 2{x_2} \Rightarrow {x_2} = 2{x_1}$$
$$\therefore$$ $$P({x_1},\sqrt {x_1^2 - 1} )$$
and $$N({x_2},0) = (2{x_1},0)$$
As tangent intersect X-axis at $$M\left( {{1 \over x},0} \right)$$.
Centroid of $$\Delta PWN = (l,m)$$
$$ \Rightarrow \left( {{{3{x_1} + {1 \over {{x_1}}}} \over 3},{{{y_1} + 0 + 0} \over 3}} \right) = (l,m)$$
$$ \Rightarrow l = {{3{x_1} + {1 \over {{x_1}}}} \over 3}$$
On differentiating w.r.t. x1, we get
$${{dl} \over {d{x_1}}} = {{3 - {1 \over {x_1^2}}} \over 3}$$
$$ \Rightarrow {{dl} \over {d{x_1}}} = 1 - {1 \over {3x_2^1}}$$, for x1 > 1
and $$m = {{\sqrt {x_1^2 - 1} } \over 3}$$
On differentiating w.r.t. x1, we get
$${{dm} \over {d{x_1}}} = {{2{x_1}} \over {2 \times 3\sqrt {x_1^2 - 1} }} = {{{x_1}} \over {3\sqrt {x_1^2 - 1} }}$$, for x1 > 1
Also, $$m = {{{y_1}} \over 3}$$
On differentiating w.r.t. y1, we get
$${{dm} \over {d{y_1}}} = {1 \over 3}$$, for y1 > 0
Let $$P\left( {{{t_1^2} \over 2},{t_1}} \right)$$ and $$Q\left( {{{t_2^2} \over 2},{t_2}} \right)$$ be two distinct points on the parabola $${y^2} = 2x$$.
The circle with PQ as diameter passes through the vertex O(0, 0) of the parabola.
Clearly, PO $$\bot$$ OQ
So, slope of PO $$\times$$ slope of OQ = $$-$$1
or, $${{{t_1} - 0} \over {{{t_1^2} \over 2} - 0}} \times {{{t_2} - 0} \over {{{t_2^2} \over 2} - 0}} = - 1$$
or, $${2 \over {{t_1}}} \times {2 \over {{t_2}}} = - 1$$
or, $${t_1}{t_2} = - 4$$
By question, area of $$\Delta OPQ = 3\sqrt 2 $$
or, $${1 \over 2}\left| {\matrix{
0 & 0 & 1 \cr
{{{t_1^2} \over 2}} & {{t_1}} & 1 \cr
{{{t_2^2} \over 2}} & {{t_2}} & 1 \cr
} } \right| = 3\sqrt 2 $$
or, $${1 \over 2}\left| {{{t_1^2{t_2}} \over 2} - {{{t_1}t_2^2} \over 2}} \right| = 3\sqrt 2 $$
or, $$\left| {{t_1}{t_2}} \right|\left| {{t_1} - {t_2}} \right| = 12\sqrt 2 $$
or, $$\left| {{t_1} + {4 \over {{t_1}}}} \right| = 3\sqrt 2 $$ [$$\because$$ $${t_1}{t_2} = - 4$$]
or, $${t_1} + {4 \over {{t_1}}} = 3\sqrt 2 $$ [$$\because$$ P lies in first quadrant]
or, $$t_1^2 - 3\sqrt 2 {t_1} + 4 = 0$$
or, $${t_1} = {{3\sqrt 2 \pm \sqrt {18 - 4 \times 1 \times 4} } \over 2}$$
$$ = {{3\sqrt 2 \pm \sqrt 2 } \over 2}$$
$$ = 2\sqrt 2 ,\sqrt 2 $$
$$\therefore$$ coordinates of $$P = (4,2\sqrt 2 )$$ or $$(1,\sqrt 2 )$$.
Therefore, (A) and (D) are correct options.