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 -

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

An ellipse intersects the hyperbola $2x^2 - 2y^2 = 1$ orthogonally. The eccentricity of the ellipse is the reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes,then:
$(A)$ Equation of ellipse is $x^2 + 2y^2 = 2$
$(B)$ The foci of ellipse are $(\pm 1, 0)$
$(C)$ Equation of ellipse is $x^2 + 2y^2 = 4$
$(D)$ The foci of ellipse are $(\pm \sqrt{2}, 0)$

If the tangent to the parabola $y^2 = x$ at a point $(\alpha, \beta)$,$(\beta > 0)$ is also a tangent to the ellipse $x^2 + 2y^2 = 1$,then $\alpha$ is equal to

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

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