Let the tangent to the curve $x^{2}+2 x-4 y+9=0$ at the point $\mathrm{P}(1,3)$ on it meet the $y$-axis at $\mathrm{A}$. Let the line passing through $\mathrm{P}$ and parallel to the line $x-3 y=6$ meet the parabola $y^{2}=4 x$ at $\mathrm{B}$. If $\mathrm{B}$ lies on the line $2 x-3 y=8$, then $(\mathrm{AB})^{2}$ is equal to ___________.

Given, equation of curve is <br/><br/>$x^2+2 x-4 y+9=0$ ..........(i) <br/><br/>Equation of tangent at $P(1,3)$ to the given curve (i) <br/><br/>$$ \begin{array}{rlrl} & x(1)+2\left(\frac{x+1}{2}\right)-4\left(\frac{y+3}{2}\right)+9 =0 \\\\ & \Rightarrow 2 x+2 x+2-4 y-12+18 =0 \\\\ &\Rightarrow 4 x-4 y+8 =0 \\\\ &\Rightarrow x-y+2 =0 \end{array} $$ <br/><br/>which is meet the $Y$-axis at $A$ <br/><br/>$\therefore A \equiv(0,2)$ <br/><br/>Equation of line passing through $P$ and parallel to $x-3 y=6$ is $x-3 y+8=0$ <br/><br/>Since, line (ii) meet the parabola $y^2=4 x$ at $B$ <br/><br/>$$ \begin{array}{lc} &\therefore y^2=4(3 y-8) \\\\ &\Rightarrow y^2-12 y+32=0 \\\\ &\Rightarrow (y-4)(y-8)=0 \end{array} $$ <br/><br/>$\therefore$ Possible co-ordinates of $B$ are $(4,4)$ and $(16,8)$. <br/><br/>Since, $(4,4)$ does not satisfies line $2 x-3 y=8$ <br/><br/>Thus, $B$ is $(16,8)$ <br/><br/>$\therefore (A B)^2=(16-0)^2+(8-2)^2=256+36=292$