Consider the circle $x^2+y^2=9$ and the parabola $y^2=8x$. They intersect at $P$ and $Q$ in the first and the fourth quadrants,respectively. Tangents to the circle at $P$ and $Q$ intersect the $x$-axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the $x$-axis at $S$.
$1.$ The ratio of the areas of the triangles $PQS$ and $PQR$ is
$(A)$ $1:\sqrt{2}$ $(B)$ $1:2$ $(C)$ $1:4$ $(D)$ $1:8$
$2.$ The radius of the circumcircle of the triangle $PRS$ is
$(A)$ $5$ $(B)$ $3\sqrt{3}$ $(C)$ $3\sqrt{2}$ $(D)$ $2\sqrt{3}$
$3.$ The radius of the incircle of the triangle $PQR$ is
$(A)$ $4$ $(B)$ $3$ $(C)$ $8/3$ $(D)$ $2$
Give the answer for questions $1, 2$ and $3.$

At the point of intersection of the rectangular hyperbola $xy = c^2$ and the parabola $y^2 = 4ax$,tangents to the rectangular hyperbola and the parabola make an angle $\theta$ and $\phi$ respectively with the $X$-axis. Then:

If the common tangents to the parabola $x^2 = 4y$ and the circle $x^2 + y^2 = 4$ intersect at the point $P$,then find the square of the slope of the line.

The locus of the midpoints of the chords of the hyperbola $x^2 - y^2 = a^2$ which are tangents to the parabola $x^2 = 4by$ will be -

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 . . . . . . .