If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is
$\frac{{2\sqrt 2 - 1}}{2}$
$\sqrt 2 - 1$
$\frac{1}{2}$
$\frac{{\sqrt 2 - 1}}{2}$
The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$
An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is
The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $4 x ^{2}+9 y ^{2}=36$
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$