Let $S$ and $S'$ be the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and $P$ be a variable point on the ellipse. The maximum area of the triangle $PSS'$ is ............. square units.

Let the line $y=mx$ and the ellipse $2x^{2}+y^{2}=1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the coordinate axes at $(-\frac{1}{3\sqrt{2}}, 0)$ and $(0, \beta)$,then $\beta$ is equal to

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

From the point $C(0, \lambda)$,two tangents are drawn to the ellipse $x^2 + 2y^2 = 4$,cutting the major axis at $A$ and $B$. If the area of $\Delta ABC$ is minimum,then the value of $\lambda$ is-

Let the line $y-x=1$ intersect the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at the points $A$ and $B$. Then the angle subtended by the line segment $AB$ at the center of the ellipse 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