Let F_1(-1, 0) and F_2(1, 0) be the foci of t | vedclass

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(C) $1.$ The correct option is $A \left(-\frac{9}{10}, 0\right)$.Equation of the ellipse: $\frac{x^2}{9}+\frac{y^2}{8}=1$. Here $a^2=9, b^2=8$. Eccent

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$A, C$

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(C) $1.$ The correct option is $A \left(-\frac{9}{10}, 0ight)$.Equation of the ellipse: $\frac{x^2}{9}+\frac{y^2}{8}=1$. Here $a^2=9, b^2=8$. Eccentricity $e = \sqrt{1-\frac{8}{9}} = \frac{1}{3}$. Foci are $(\pm ae, 0) = (\pm 1, 0)$.Equation of the parabola with vertex $(0,0)$ and focus $F_2(1,0)$ is $y^2 = 4x$.Substituting $y^2=4x$ into the ellipse equation: $\frac{x^2}{9} + \frac{4x}{8} = 1 \Rightarrow 2x^2 + 9x - 18 = 0 \Rightarrow (2x-3)(x+6)=0$. Since $x>0$,$x=\frac{3}{2}$. Then $y^2 = 4(\frac{3}{2}) = 6$,so $y = \pm \sqrt{6}$. Thus $M = (\frac{3}{2}, \sqrt{6})$ and $N = (\frac{3}{2}, -\sqrt{6})$.In $	riangle F_1 M N$,the altitude from $M$ to $F_1 N$ has slope $m_1 = \frac{\sqrt{6} - 0}{3/2 - (-1)} = \frac{\sqrt{6}}{5/2} = \frac{2\sqrt{6}}{5}$. The slope of $F_1 N$ is $\frac{-\sqrt{6}-0}{3/2 - (-1)} = \frac{-\sqrt{6}}{5/2} = -\frac{2\sqrt{6}}{5}$. The slope of the altitude from $M$ is $m_{alt} = -\frac{1}{-2\sqrt{6}/5} = \frac{5}{2\sqrt{6}}$.Equation of altitude from $M$: $y - \sqrt{6} = \frac{5}{2\sqrt{6}}(x - \frac{3}{2})$. Since the orthocentre lies on the $x$-axis (as $F_1 M N$ is isosceles with $F_1 M = F_1 N$),set $y=0$: $-\sqrt{6} = \frac{5}{2\sqrt{6}}(x - \frac{3}{2}) \Rightarrow -12 = 5x - \frac{15}{2} \Rightarrow 5x = -12 + 7.5 = -4.5 \Rightarrow x = -\frac{9}{10}$.$2.$ The tangent to the ellipse at $M(\frac{3}{2}, \sqrt{6})$ is $\frac{x(3/2)}{9} + \frac{y\sqrt{6}}{8} = 1 \Rightarrow \frac{x}{6} + \frac{y\sqrt{6}}{8} = 1$. For $y=0$,$x=6$,so $R(6,0)$.The normal to the parabola $y^2=4x$ at $M(\frac{3}{2}, \sqrt{6})$: slope of tangent is $\frac{dy}{dx} = \frac{2}{y} = \frac{2}{\sqrt{6}}$. Slope of normal is $-\frac{\sqrt{6}}{2}$. Equation: $y - \sqrt{6} = -\frac{\sqrt{6}}{2}(x - \frac{3}{2})$. For $y=0$,$\sqrt{6} = \frac{\sqrt{6}}{2}(x - \frac{3}{2}) \Rightarrow 2 = x - \frac{3}{2} \Rightarrow x = \frac{7}{2}$. So $Q(\frac{7}{2}, 0)$.Area $	riangle MQR = \frac{1}{2} 	imes |x_R - x_Q| 	imes y_M = \frac{1}{2} 	imes (6 - 3.5) 	imes \sqrt{6} = \frac{1}{2} 	imes 2.5 	imes \sqrt{6} = \frac{5\sqrt{6}}{4}$.Area quadrilateral $MF_1NF_2 = \frac{1}{2} 	imes |F_1 F_2| 	imes |y_M - y_N| = \frac{1}{2} 	imes 2 	imes 2\sqrt{6} = 2\sqrt{6}$.Ratio $= \frac{5\sqrt{6}/4}{2\sqrt{6}} = \frac{5}{8}$.

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