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 $b^2 = \dots$

Let the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the reciprocal of the eccentricity of the hyperbola $2x^2 - 2y^2 = 1$. If the ellipse intersects the hyperbola at right angles,then the square of the length of the latus-rectum of the ellipse is $................$.

If $A = \{(x, y) : x^2 + y^2 = 25\}$ and $B = \{(x, y) : x^2 + 9y^2 = 144\}$,then $A \cap B$ contains

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then,the value of $b^2$ is

If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$,then $12\; m^{2}$ is equal to

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