Find the coordinates of the foci of the ellipse $25(x + 1)^2 + 9(y + 2)^2 = 225$.

If the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ are $\alpha$ and $\beta$,then $\frac{\cot(\alpha/2)}{\tan(\beta/2)}=$

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

Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a < b$,pass through the point $(4, 3)$ and have eccentricity $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is:

Let the line $2x + 3y - k = 0, k > 0$,intersect the $x$-axis and $y$-axis at the points $A$ and $B$,respectively. If the equation of the circle having the line segment $AB$ as a diameter is $x^2 + y^2 - 3x - 2y = 0$ and the length of the latus rectum of the ellipse $x^2 + 9y^2 = k^2$ is $\frac{m}{n}$,where $m$ and $n$ are coprime,then $2m + n$ is equal to

An ellipse has its center at $(1, -2)$,one focus at $(3, -2)$,and one vertex at $(5, -2)$. Then the length of its latus rectum is :