Let $F_1\left(x_1, 0\right)$ and $F_2\left(x_2, 0\right)$, for $x_1<0$ and $x_2>0$, be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.

($1$)The orthocentre of the triangle $F_1 M N$ is

($A$) $\left(-\frac{9}{10}, 0\right)$   ($B$) $\left(\frac{2}{3}, 0\right)$    ($C$) $\left(\frac{9}{10}, 0\right)$    ($D$) $\left(\frac{2}{3}, \sqrt{6}\right)$

($2$) If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F_{\mathrm{I}} N F_2$ is

($A$) $3: 4$     ($B$) $4: 5$     ($C$) $5: 8$     ($D$) $2: 3$

Givan the answer qestion ($1$) and ($2$)

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{4 a^{2}}=1$ and the co-ordinate axis is $kab,$ then $\mathrm{k}$ is equal to ..... .

The line, $ lx + my + n = 0$  will cut the ellipse $\frac{{{x^2}}}{{{a^2}}}$ $+$ $\frac{{{y^2}}}{{{b^2}}}$ $= 1 $ in points whose eccentric angles differ by $\pi /2$  if :

If the eccentricity of an ellipse be $5/8$ and the distance between its foci be $10$, then its latus rectum is

If the line $y = mx + c$touches the ellipse $\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$, then $c = $

A ray of light through $(2,1)$ is reflected at a point $P$ on the $y$ - axis and then passes through the point $(5,3)$. If this reflected ray is the directrix of an ellipse with eccentrieity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be :