Step-by-step Solution

1. Find the intersection points of the circle and the parabola.

The given equations are:

• Circle: ${x^2} + {y^2} = 9$ (1)
• Parabola: ${y^2} = 8x$ (2)

To find the intersection points, substitute equation (2) into equation (1): ${x^2} + 8x = 9$ ${x^2} + 8x - 9 = 0$ Factoring the quadratic equation: $(x+9)(x-1) = 0$ This gives two possible values for x: $x = 1$ or $x = -9$.

• If $x = -9$, from equation (2), ${y^2} = 8(-9) = -72$. This is not possible for real values of y.
• If $x = 1$, from equation (2), ${y^2} = 8(1) = 8$. This gives $y = \pm \sqrt{8} = \pm 2\sqrt{2}$.

The intersection points are $(1, 2\sqrt{2})$ and $(1, -2\sqrt{2})$.

• Point P is in the first quadrant, so $P = (1, 2\sqrt{2})$.
• Point Q is in the fourth quadrant, so $Q = (1, -2\sqrt{2})$.

2. Find the point R (intersection of tangents to the circle with the x-axis).

The equation of the tangent to the circle ${x^2} + {y^2} = {r^2}$ at a point $(x_1, y_1)$ is $x{x_1} + y{y_1} = {r^2}$. Here, ${r^2} = 9$.

• Tangent to the circle at $P(1, 2\sqrt{2})$: $x(1) + y(2\sqrt{2}) = 9 \implies x + 2\sqrt{2}y = 9$

To find the intersection point R with the x-axis, set $y = 0$: $x + 2\sqrt{2}(0) = 9 \implies x = 9$. So, the point R is $(9, 0)$. Note that the tangent at Q is $x - 2\sqrt{2}y = 9$, which also intersects the x-axis at $(9,0)$.

3. Find the point S (intersection of tangents to the parabola with the x-axis).

The equation of the tangent to the parabola ${y^2} = 4ax$ at a point $(x_1, y_1)$ is $y{y_1} = 2a(x + {x_1})$. For the parabola ${y^2} = 8x$, we have $4a = 8$, so $a = 2$. The tangent equation is $y{y_1} = 4(x + {x_1})$.

• Tangent to the parabola at $P(1, 2\sqrt{2})$: $y(2\sqrt{2}) = 4(x + 1)$

To find the intersection point S with the x-axis, set $y = 0$: $0 = 4(x + 1) \implies x = -1$. So, the point S is $(-1, 0)$. The tangent at Q, $y(-2\sqrt{2}) = 4(x+1)$, also intersects the x-axis at $(-1,0)$. A known property for a parabola $y^2 = 4ax$ is that the tangent at $(x_1, y_1)$ meets the x-axis at $(-x_1, 0)$. Here $x_1=1$, so the intersection is at $(-1,0)$, which confirms our result for S.

4. Calculate the ratio of the areas of $\triangle PQS$ and $\triangle PQR$.

Both triangles, $\triangle PQS$ and $\triangle PQR$, share the same base PQ. The base PQ is a vertical line segment along the line $x=1$. The length of the base PQ is the distance between $P(1, 2\sqrt{2})$ and $Q(1, -2\sqrt{2})$, which is $2\sqrt{2} - (-2\sqrt{2}) = 4\sqrt{2}$.

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$. Since the base PQ is common, the ratio of the areas is equal to the ratio of their corresponding heights.

• For $\triangle PQS$, the vertex is $S(-1, 0)$. The base PQ lies on the line $x=1$. The height of $\triangle PQS$ is the perpendicular distance from S to the line $x=1$. Height$_S = |x_P - x_S| = |1 - (-1)| = |2| = 2$.

• For $\triangle PQR$, the vertex is $R(9, 0)$. The base PQ lies on the line $x=1$. The height of $\triangle PQR$ is the perpendicular distance from R to the line $x=1$. Height$_R = |x_R - x_P| = |9 - 1| = |8| = 8$.

Now, let's find the ratio of the areas: $\frac{\text{Area}(\triangle PQS)}{\text{Area}(\triangle PQR)} = \frac{\frac{1}{2} \times |PQ| \times \text{Height}_S}{\frac{1}{2} \times |PQ| \times \text{Height}_R} = \frac{\text{Height}_S}{\text{Height}_R}$ $\frac{\text{Area}(\triangle PQS)}{\text{Area}(\triangle PQR)} = \frac{2}{8} = \frac{1}{4}$

The ratio of the areas of the triangles PQS and PQR is 1 : 4.