Let $P(x_1, y_1)$ and $Q(x_2, y_2)$,with $y_1 < 0$ and $y_2 < 0$,be the endpoints of the latus rectum of the ellipse $x^2 + 4y^2 = 4$. The equations of the parabolas with latus rectum $PQ$ are:
$(A) x^2 + 2\sqrt{3}y = 3 + \sqrt{3}$
$(B) x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$
$(C) x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$
$(D) x^2 - 2\sqrt{3}y = 3 - \sqrt{3}$

The slope of a common tangent to the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ and the circle $x^2+y^2=16$ is

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be an ellipse,whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ 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
$(A)$ $3: 4$ $(B)$ $4: 5$ $(C)$ $5: 8$ $(D)$ $2: 3$

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=8ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $e^4$ is equal to

If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles,then the value of $b$ is