The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{\pi}{6}$ sq. units. Which of the following is a possible equation of the ellipse?

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(b > a)$ is an ellipse with eccentricity $e = \frac{1}{\sqrt{2}}$. If the angle of intersection between the ellipse and the parabola $y^2 = 4ax$ is $\theta$,then the coordinates of the point corresponding to $\theta$ on the ellipse are:

Let $A_1$ be the area of the given ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let $A_2$ be the area of the region bounded by the curve which is the locus of the midpoint of the line segment joining the focus of the ellipse and a point $P$ on the given ellipse. Then $A_1 : A_2$ is equal to:

If the equations $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$ for $0 \leq \theta < 2 \pi$ represent an ellipse,then the point of intersection of the normal drawn at $P(\theta = \pi/4)$ to this ellipse and its major axis is:

If $\alpha, \beta$ are the two real roots of the $4^{th}$ roots of unity and $\gamma, \delta$ are the other two roots,then the sum of the eccentricities of the conics $|z-\alpha|+|z-\beta|=4$ and $|z-\gamma|+|z-\delta|=6$ is

If the point $P$ on the curve $4x^{2} + 5y^{2} = 20$ is farthest from the point $Q(0, -4)$,then $PQ^{2}$ is equal to: