$1.$ The correct option is $A \left(-\frac{9}{10}, 0\right)$

$Image$

Equation of the ellipse:

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

$a=3, b=2 \sqrt{2}$

Hence, eccentricity is

$e^2=1-\frac{b^2}{a^2}=\frac{1}{9}$

$\Rightarrow e=\frac{1}{3}$

So, foci of ellipse are ( $\pm 1,0$ )

Equation of the parabola having vertex $(0,0)$ and focus $(1,0)$ is

$y^2=4 x$

From equations $(1)$ and $(2)$, we get

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

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

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

$\Rightarrow x=\frac{3}{2} \quad[\because x=-6 \rightarrow \text { not possible }]$

$\therefore M \equiv\left(\frac{3}{2} \cdot \sqrt{6}\right), \quad N \equiv\left(\frac{3}{2}-\sqrt{6}\right)$

Equation of altitude from vertex $M\left(\frac{3}{2}, \sqrt{6}\right)$ is

$(y-\sqrt{6})=\frac{5}{2 \sqrt{6}}\left(x-\frac{3}{2}\right)$

$\because x$-axis is altitude drawn through vertex $F_1$

$\therefore$ orthocentre lies on the $x$-axis,

$\therefore 0-\sqrt{6}=\frac{5}{2 \sqrt{6}}\left(x-\frac{3}{2}\right)$

$\Rightarrow x =-\frac{9}{10}$

Hence, the orthocentre of $\triangle F_1 M N$ is

$\left(-\frac{9}{10} 0\right)$

$2.$  $Image$ Equation of tangent at point $M\left(\frac{3}{2}, \sqrt{6}\right)$ to the ellipse is $\frac{x\left(\frac{3}{2}\right)}{9}+\frac{y \sqrt{6}}{8}=1$ Put $y=0 \Rightarrow R$ is $(6,0) \quad[\because R$ lies on $x$-axis $)$

Equation of the normal to the parabola at point $M\left(\frac{3}{2} \sqrt{6}\right)$ is

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

Put $y=0 \Rightarrow Q$ is $\left(\frac{7}{2}, 0\right) \quad[\because Q$ lies on $x$-axis $)$

Area of $\triangle M Q R=\frac{1}{2} \times \sqrt{6} \times \frac{5}{2}=\frac{5 \sqrt{6}}{4}$

Area of quadrilateral $MF _1 NF _2$

$=$ Area of $\triangle M F_1 F_2+$ Area of $\triangle N F_1 F_2$

$=\sqrt{6}+\sqrt{6}=2 \sqrt{6}$

$\therefore$ Area of $\triangle M Q R$ : Area of quadrilateral $M F_1 N F_2=5: 8$