P is point on $y^2=12x$ such that its focal distance is 6. Tangent drawn at P intersect tangent at vertex at point T (y > 0). Point R is on x-axis inside parabola such that SR = 6 (S is focus). M is a point on parabola such that tangents at M & P meet at point N such that SN = 5, then which of the following is/are TRUE?

$$\begin{gathered} As SP=6=3t^2+3\Rightarrow t=1 \Rightarrow P=(3,6) \\ Tangent at P: y = 6 (x + 3) \\ \Rightarrow x-y+3=0 \Rightarrow T=(0,3)\Rightarrow ST=3\sqrt{2} \\ Also Q=(-3,0)\Rightarrow QS=6 \end{gathered}$$

As given SR = 6  $\Rightarrow R(9,0)$

As we know if tangent and normal at any point P cuts axis at Q & R then SP = SQ = SR
According to figure PR is normal
Now,
Area of PRST = Area of $\Delta PQR$ - Area of  $\Delta TQS$
$=\frac{1}{2}\times PS\times QR-\frac{1}{2}\times OT\times QS=\frac{1}{2}\times 6\times 12-\frac{1}{2}\times 3\times 6=27$
Also as we know if tangent at P & M meet at point N, then
$$\begin{gathered} {\left(SN\right)}^2=SP\times SM \\ 25=6\times SM \end{gathered}$$

$\Rightarrow SM=\frac{25}{6}=3t^2+3\Rightarrow t^2=\frac{7}{18}\Rightarrow t=\pm \frac{\sqrt{7}}{3\sqrt{2}}$

Ordinate of M = 2at

$=\mathrm{2.3.}(\pm \frac{\sqrt{7}}{3\sqrt{2}})=\pm \sqrt{14}$

sum of squares = 14 + 14 = 28