If the eccentricity of the ellipse $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$ is $\frac{1}{\sqrt{6}}$,find the length of the latus rectum of the ellipse.

Let the point $L$ lying in the first quadrant be one end of a latus rectum of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $P$ and $Q$ be the points where the normal drawn at $L$ to this given ellipse meets the major axis and the minor axis. Then the distance between $P$ and $Q$ is

$L_1^{\prime}$ is the end of a latus rectum of the ellipse $3x^2 + 4y^2 = 12$ which is lying in the third quadrant. If the normal drawn at $L_1^{\prime}$ to this ellipse intersects the ellipse again at the point $P(a, b)$,then $a =$

For the ellipse $25x^2 + 9y^2 - 150x - 90y + 225 = 0$,the eccentricity $e$ is: (in $/5$)

Which of the following points lies on the locus of the foot of the perpendicular drawn from any of the foci of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ to any of its tangents?

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.