If the distance between the directrices is thrice the distance between the foci,then the eccentricity of the ellipse is

Let $A(\alpha, 0)$ and $B(0, \beta)$ be the points on the line $5x + 7y = 50$. Let the point $P$ divide the line segment $AB$ internally in the ratio $7:3$. Let $3x - 25 = 0$ be a directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the corresponding focus be $S$. If from $S$,the perpendicular on the $x$-axis passes through $P$,then the length of the latus rectum of $E$ is equal to

If $m$ is the length of the latus rectum and $n$ is the length of the major axis of the ellipse $25x^2+16y^2-150x-64y-111=0$,then the ordered pair $(m, n) =$

In an ellipse with its centre at the origin,if the difference between the lengths of the major axis and the minor axis is $10$ and one of the foci is at $(0, 5\sqrt{3})$,then the length of its latus rectum is:

The length of the latus rectum of an ellipse is $6$ units and the distance between a focus and its nearest vertex on the major axis is $\frac{5}{3}$ units. If $e$ is the eccentricity of this ellipse,then $e$ satisfies the equation:

The value of $\lambda$,for which the line $2x - \frac{8}{3}\lambda y = -3$ is a normal to the conic $x^2 + \frac{y^2}{4} = 1$ is