F₁(x, 0) and F₂(x₂, 0) are the foci of the ellipse:

$${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$

Therefore, a² = 9 and b² = 8.

$${b^2} = {a^2}(1 - {e^2})$$

$$1 - {e^2} = {8 \over 9} \Rightarrow {e^2} = 1 - {8 \over 9} = {1 \over 9} \Rightarrow e = {1 \over 3}$$

The focus is

$${F_1}\left( { - 3 \times {1 \over 3},0} \right)$$ and $${F_2}\left( {3 \times {1 \over 3},0} \right)$$

That is, F₁($$-$$1, 0) and F₂(1, 0).

The equation of parabola is

$${y^2} = 4(O{F_2})x$$

$${y^2} = 4x(O{F_2} = 1)$$

The point of intersection of ellipse and parabola is

$${{{x^2}} \over 9} + {{4x} \over 8} = 1 \Rightarrow {{{x^2}} \over 9} + {x \over 2} = 1$$

$$ \Rightarrow 2{x^2} + 9x - 18 = 0$$

$$ \Rightarrow 2{x^2} + 12x - 3x - 18 = 0$$

$$ \Rightarrow 2x(x + 6) - 3(x + 6) = 0$$

$$ \Rightarrow x = {3 \over 2}$$ (x $$-$$6 is rejected)

Now, $${y^2}(4){3 \over 2} = 6$$

$$y = \pm \sqrt 6 $$

That is, the points M and N are, respectively, $$M\left( {{3 \over 2},\sqrt 6 } \right)$$ and $$N\left( {{3 \over 2}, - \sqrt 6 } \right)$$.

Let the orthocenter be (h, k).

The slope of $$OM = {{k - \sqrt 6 } \over {h - (3/2)}}$$

The slope of $$ON = {{\sqrt 6 } \over { - 1 - (3/2)}} = {{ - 2\sqrt 6 } \over 5}$$

Now, $$\left( {{{k - \sqrt 6 } \over {h - (3/2)}}} \right)\left( {{{ - 2\sqrt 6 } \over 5}} \right) = - 1$$

$$2\sqrt 6 k - 12 = 5h - {{15} \over 2}$$

$$5h - 2\sqrt 6 k = {{15} \over 2} - 12 = {{ - 9} \over 2}$$

The slope of $$ON = {{k + \sqrt 6 } \over {h - (3/2)}}$$

The slope of $${F_1}M = {{\sqrt 6 } \over {1 + (3/2)}} = {{2\sqrt 6 } \over 5}$$

$${{k + \sqrt 6 } \over {h - (3/2)}} \times {{2\sqrt 6 } \over 5} = - 1$$

$$2\sqrt 6 k + 12 = - 5h + {{15} \over 2}$$

$$5h + 2\sqrt 6 k = {{15} \over 2} - 12 = {{ - 9} \over 2}$$

$$5h + 2\sqrt 6 k = {{ - 9} \over 2}$$ ....... (1)

$$5h - 2\sqrt 6 k = {{ - 9} \over 2}$$ ........ (2)

Solving Eqs. (1) and (2), we get

$$10h = - 9 \Rightarrow h = {{ - 9} \over {10}}$$ and k = 0

Hence, the orthocentre of the triangle F₁MN is $$\left( {{{ - 9} \over {10}},0} \right)$$.

Equation of tangent at M(3/2, $$\sqrt6$$) to $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$ is

$${3 \over 2}.{x \over 9} + \sqrt 6 .{y \over 8} = 1$$ ....... (i)

which intersect X-axis at (6, 0).

Also, equation of tangent at N(3/2, $$-$$$$\sqrt6$$) is

$${3 \over 2}.{x \over 9} - \sqrt 6 .{y \over 8} = 1$$ ....... (ii)

Eqs. (i) and (ii) intersect on X-axis at R(6, 0). ........ (iii)

Also, normal at $$M(3/2,\sqrt 6 )$$ is

$$y - \sqrt 6 = {{ - \sqrt 6 } \over 2}\left( {x - {3 \over 2}} \right)$$

On solving with y = 0, we get Q(7/2, 0) ....... (iv)

The area of MQR is

$$\left| {{1 \over 2}} \right|\left| {\matrix{ {3/2} & {\sqrt 6 } & 1 \cr 6 & 0 & 1 \cr {7/2} & 0 & 1 \cr } } \right| = \left| {{{\sqrt 6 } \over 2}\left( {6 - {7 \over 2}} \right)} \right| = {{5\sqrt 6 } \over 4}$$

The area of the quadrilateral MF₁NF₂ is

$$2(\Delta {m_1}{F_1}{F_2}) = 2\sqrt 6 $$

and the required ratio is

$${{5\sqrt 6 } \over {4\,.\,2\sqrt 6 }} = {5 \over 8}$$