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.