The ratio of the areas of the triangle $PQS$ and $PQR$ is :

Equation of parabola is $y^2=8x$

Equation of circle is $x^2+y^2=9$

To find point of intersection of both, put equation of parabola in equation of circle.

$\therefore x^2+8x=9$

$\therefore x^{ 2 }+8x-9=0$

$\therefore x^{ 2 }+9x-x-9=0$

$\therefore x\left( x+9 \right) -1\left( x+9 \right) =0$

$\therefore \left( x+9 \right) \left( x-1 \right) =0$

$\therefore x=-9$ and $x=1$

But, $x\neq -9$ as $y^2$ cannot be negative

$\therefore x=1$

From equation of parabola,

${ y }^{ 2 }=8$

$\therefore { y }=\pm 2\sqrt { 2 }$

Thus, points of intersection are $P\left( 1,2\sqrt { 2 } \right)$ and $Q\left( 1,-2\sqrt { 2 } \right)$

Consider tangent to the circle which intersects x axis at point R.

Equation of tangent is,

$x{ x }_{ 1 }+y{ y }_{ 1 }=9$

$\therefore x\left( 1 \right) +y\left( 2\sqrt { 2 } \right) =9$

$\therefore x+2\sqrt { 2 } y=9$

When $y=0$, $x=9$

Thus, coordinates of point R are $R\left( 9,0 \right)$

Consider tangent to the parabola which intersects x axis at point S.

Equation of tangent is,

$y{ y }_{ 1 }=4a\left( x+{ x }_{ 1 } \right)$

$\therefore y\left( 2\sqrt { 2 } \right) =8\left( x+1 \right)$

When

$y=0$, $x=-1$

Thus, coordinates of point S are $R\left( -1,0 \right)$

Area of triangle PQS is,

$A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times PQ\times SM$

$\therefore A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 2$

$\therefore A\left( \triangle PQS \right) =4\sqrt { 2 }$

Area of triangle PQR is,

$A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times PQ\times RM$

$\therefore A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 8$

$\therefore A\left( \triangle PQR \right) =16\sqrt { 2 }$

$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =\dfrac { 4\sqrt { 2 } }{ 16\sqrt { 2 } }$

$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =1:4$