Step-by-step Solution:

1. Analyze the Ellipse

The equation of the ellipse is given by ${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$.

• Comparing this with the standard form ${{{x^2}} \over a^2} + {{{y^2}} \over b^2} = 1$, we have $a^2 = 9$ and $b^2 = 8$. Thus, $a=3$ and $b=\sqrt{8} = 2\sqrt{2}$.
• The eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$. $8 = 9(1 - e^2) \implies {8 \over 9} = 1 - e^2 \implies e^2 = 1 - {8 \over 9} = {1 \over 9} \implies e = {1 \over 3}$
• The foci are at $(\pm ae, 0)$. $ae = 3 \times {1 \over 3} = 1$
• So, the foci are $(\pm 1, 0)$. Since it's given that $x_1 < 0$ and $x_2 > 0$, the foci are $F_1(-1, 0)$ and $F_2(1, 0)$.

2. Analyze the Parabola

• The parabola has its vertex at the origin $(0, 0)$ and its focus at $F_2(1, 0)$.
• The standard equation of a parabola with vertex at $(0,0)$ and focus at $(a', 0)$ is $y^2 = 4a'x$.
• In this case, $a' = 1$. So, the equation of the parabola is $y^2 = 4x$.

3. Find Intersection Points M and N

To find the intersection points of the ellipse and the parabola, we solve their equations simultaneously:

• Ellipse: ${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$
• Parabola: $y^2 = 4x$ Substitute $y^2 = 4x$ into the ellipse equation: ${{{x^2}} \over 9} + {{4x} \over 8} = 1$ ${{{x^2}} \over 9} + {{x} \over 2} = 1$ Multiply by 18 to clear the denominators: $2x^2 + 9x = 18$ $2x^2 + 9x - 18 = 0$ Factor the quadratic equation: $2x^2 + 12x - 3x - 18 = 0$ $2x(x + 6) - 3(x + 6) = 0$ $(2x - 3)(x + 6) = 0$ This gives two possible values for $x$: $x = 3/2$ or $x = -6$.
• If $x = -6$, then $y^2 = 4(-6) = -24$, which is not possible for real $y$.
• If $x = 3/2$, then $y^2 = 4(3/2) = 6$, so $y = \pm \sqrt{6}$.

The intersection points are $(3/2, \sqrt{6})$ and $(3/2, -\sqrt{6})$.

• Point $M$ is in the first quadrant, so $M = (3/2, \sqrt{6})$.
• Point $N$ is in the fourth quadrant, so $N = (3/2, -\sqrt{6})$.

4. Find Point R

Point $R$ is the intersection of the tangents to the ellipse at $M$ and $N$. The equation of the tangent to the ellipse at $(x_0, y_0)$ is ${xx_0 \over a^2} + {yy_0 \over b^2} = 1$.

• For points of the form $(x_0, y_0)$ and $(x_0, -y_0)$, the tangents intersect on the x-axis at the point $(\frac{a^2}{x_0}, 0)$.
• Here, $x_0 = 3/2$ and $a^2 = 9$. So, the x-coordinate of R is: $x_R = {9 \over {3/2}} = 9 \times {2 \over 3} = 6$
• Thus, the point $R$ is $(6, 0)$.

5. Find Point Q

Point $Q$ is the intersection of the normal to the parabola at $M$ with the x-axis. The parabola is $y^2 = 4x$.

• Differentiating with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.
• At point $M(3/2, \sqrt{6})$, the slope of the tangent is $m_T = \frac{2}{\sqrt{6}}$.
• The slope of the normal is $m_N = -\frac{1}{m_T} = -\frac{\sqrt{6}}{2}$.
• The equation of the normal at $M$ is: $y - y_M = m_N(x - x_M)$ $y - \sqrt{6} = -\frac{\sqrt{6}}{2} \left(x - \frac{3}{2}\right)$
• To find where it meets the x-axis, set $y=0$: $0 - \sqrt{6} = -\frac{\sqrt{6}}{2} \left(x_Q - \frac{3}{2}\right)$ $1 = \frac{1}{2} \left(x_Q - \frac{3}{2}\right)$ $2 = x_Q - \frac{3}{2}$ $x_Q = 2 + \frac{3}{2} = \frac{7}{2}$
• Thus, the point $Q$ is $(7/2, 0)$.

6. Calculate Area of Triangle MQR

The vertices of the triangle are $M(3/2, \sqrt{6})$, $Q(7/2, 0)$, and $R(6, 0)$.

• The base of the triangle lies on the x-axis, from $Q$ to $R$. Length of base $QR = |x_R - x_Q| = |6 - 7/2| = |12/2 - 7/2| = 5/2$.
• The height of the triangle is the y-coordinate of $M$, which is $\sqrt{6}$.
• Area of $\triangle MQR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{5}{2} \times \sqrt{6} = \frac{5\sqrt{6}}{4}$.

7. Calculate Area of Quadrilateral MF1NF2

The vertices are $M(3/2, \sqrt{6})$, $F_1(-1, 0)$, $N(3/2, -\sqrt{6})$, and $F_2(1, 0)$.

• This quadrilateral is a kite, with diagonals $F_1F_2$ and $MN$.
• The length of diagonal $F_1F_2$ is $|1 - (-1)| = 2$.
• The length of diagonal $MN$ is $|\sqrt{6} - (-\sqrt{6})| = 2\sqrt{6}$.
• Area of a kite is $\frac{1}{2} d_1 d_2$.
• Area of quadrilateral $M{F_1}N{F_2} = \frac{1}{2} \times (2) \times (2\sqrt{6}) = 2\sqrt{6}$.

8. Find the Ratio

The required ratio is: $\frac{\text{Area of } \triangle MQR}{\text{Area of quadrilateral } M{F_1}N{F_2}} = \frac{5\sqrt{6}/4}{2\sqrt{6}} = \frac{5\sqrt{6}}{4} \times \frac{1}{2\sqrt{6}} = \frac{5}{8}$

• The ratio is $5:8$.

Comparing this with the options, the correct option is C.