The normal at $\left( 2, \frac{3}{2} \right)$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ touches a parabola,whose equation is

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in R$. If the set of all possible values of $q$,such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$,be the interval $(\alpha, \beta]$,then $4(\alpha + \beta)$ equals :

The ellipse $4x^2 + 9y^2 = 36$ and the hyperbola $4x^2 - y^2 = 4$ have the same foci and they intersect at right angles. Then,the equation of the circle passing through the points of intersection of the two conics is:

Let $F_1(-1, 0)$ and $F_2(1, 0)$ be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having its 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 the area of the triangle $M Q R$ to the area of the quadrilateral $M F_1 N F_2$ is
$(A)$ $3: 4$ $(B)$ $4: 5$ $(C)$ $5: 8$ $(D)$ $2: 3$

For some $\theta \in (0, \frac{\pi}{2})$,let the eccentricity and the length of the latus rectum of the hyperbola $x^{2} - y^{2} \sec^{2} \theta = 8$ be $e_{1}$ and $l_{1}$,respectively,and let the eccentricity and the length of the latus rectum of the ellipse $x^{2} \sec^{2} \theta + y^{2} = 6$ be $e_{2}$ and $l_{2}$,respectively. If $e_{1}^{2} = e_{2}^{2}(\sec^{2} \theta + 1)$,then $(\frac{l_{1}l_{2}}{e_{1}e_{2}}) \tan^{2} \theta$ is equal to . . . . . . .

The area of the quadrilateral formed by the foci of the hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ is