If $e_1$,$e_2$,and $e_3$ are eccentricities of the conics $y = x^2 - x + 3$,$\frac{x^2}{a^2} + \frac{y^2}{3a^4} = 1$,and $a^2x^2 - 3a^4y^2 = 1$ respectively,then which of the following is correct? (where $a > 1$)

If for $\theta \in \left[-\frac{\pi}{3}, 0\right]$,the points $(x, y) = \left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma=0$,then $\alpha^2+\beta^2+\gamma^2$ is equal to:

The angle between the curves $x^2-y^2=4$ and $x^2+y^2=4\sqrt{2}$ is

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ are the same as the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2 = \dots$

The equations of the common tangents to the two hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are:

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