Let  $C_1:y^2=4x;C_2:{(x-4)}^2+y^2=8$ be curves in the cartesian plane. Through point P on $C_1$ draw line $\mathcal{l}$ with ${45}^{\circ }$ inclination that intersects $C_2$ at distinct points Q and R. The range of  $\left|PQ\right|.\left|PR\right|$ is (where $\left|PQ\right|$ denotes distance between the points P and Q)

Let  $P(t^2,2t)$
Equation of line $\mathcal{l}$ is  $(y-2t)=\left(1\right)(x-t^2)$$\Rightarrow y=x+2t-t^2$  
 (M is centre of the circle)
$={(t^2-4)}^2+{(2t-0)}^2-8=t^4-4t^2+8={(t^2-2)}^2+4$

If t = 0 then the line $\mathcal{l}$  is y = x, it touches the circle  ${(x-4)}^2+y^2=8$
To find range of t, use  $r>d_{\perp }$
$\left|\frac{4+2t-t^2}{\sqrt{2}}\right|<2\sqrt{2} \Rightarrow |4+2t-t^2|<4$

$-4<4+2t-t^2<4 \Rightarrow t^2-2t-8<0 t^2-2t>0$

$$\begin{gathered} {(t-1)}^2-9<0 t(t-2)>0 \\ (t-2)(t+4)<0t\in (-\infty ,0)\cup (2,\infty ) \_\_\left(2\right) \end{gathered}$$

$t\in (-2,4) \_\_\_\_\_\_\_\left(1\right)$

From (1) and (2)   $\Rightarrow t\in (-2,0)\cup (2,4)$

$\left|PQ\right|\left|PR\right|={(t^2-2)}^2+4\in [4,8)\cup (8,200)$