If the distance between the two foci of an ellipse is equal to its minor axis,then the eccentricity of the ellipse is:

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is:

The equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is

Tangents are drawn to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at all the four ends of its latus rectum. The area (in sq units) of the quadrilateral formed by these tangents is

$A$ point on the curve $x=3 \cos \theta, y=2 \sin \theta$ at which the tangent is perpendicular to the $X$-axis is