The tangents at the extremities of the latus rectum of the ellipse $3x^2 + 4y^2 = 12$ form a rhombus of area (in $sq. \ units$):

If the product of the lengths of the perpendiculars drawn from the foci to the tangent $y = \frac{-3}{4}x + 3\sqrt{2}$ of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $9$,then the eccentricity of that ellipse is

The eccentric angle of the end point of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

If the line $\alpha x+4y=\sqrt{7}$, where $\alpha \in R$, touches the ellipse $3x^{2}+4y^{2}=1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is:

Two sets $A$ and $B$ are defined as follows:
$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$
$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$
Then:

If the distance between the foci of an ellipse is half the length of its latus rectum,then the eccentricity of the ellipse is