If the normal drawn at the point $(2, -1)$ to the ellipse $x^2 + 4y^2 = 8$ meets the ellipse again at $(a, b)$,then $17a =$

If $S$ and $S^{\prime}$ are the foci of an ellipse $\frac{x^2}{169}+\frac{y^2}{144}=1$ and the point $B$ lying on the positive $Y$-axis is one end of its minor axis,then the incenter of the triangle $SBS^{\prime}$ is

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:

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, A < B$ have the same eccentricity $e = \frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$,and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$,then the area of the quadrilateral $ABCD$ equals:

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be equal to the length of its semi-major axis. If the radius of its director circle is $\sqrt{3}$ and $e$ is its eccentricity,then the length of its latus rectum is

$A$ wall is inclined to the floor at an angle of $135^{\circ}$. $A$ ladder of length $l$ is resting on the wall. As the ladder slides down,its mid-point traces an arc of an ellipse. Then,the area of the ellipse is