If the centre, one of the foci and semi-major axis of an ellipse be $(0, 0), (0, 3)$ and $5$ then its equation is

The equation of the ellipse whose centre is $(2, -3)$, one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is

The equation of the ellipse whose foci are $( \pm 5,\;0)$ and one of its directrix is $5x = 36$, is

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 ($2$)

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1 \,\,and\,\,\left| {b - 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \le 36} \right\}$ then : . . . . .

Let $PQ$ be a focal chord of the parabola $y^{2}=4 x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^{2}}$ is equal to