If the eccentricity of an ellipse be $5/8$ and the distance between its foci be $10$, then its latus rectum is

Let $\mathrm{A}(\alpha, 0)$ and $\mathrm{B}(0, \beta)$ be the points on the line $5 x+7 y=50$. Let the point $P$ divide the line segment $A B$ internally in the ratio $7: 3$. Let $3 x-$ $25=0$ 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 $\mathrm{x}$-axis passes through $\mathrm{P}$, then the length of the latus rectum of $\mathrm{E}$ is equal to

The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $P Q$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

If $\theta $ and $\phi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then $\theta - \phi $ is equal to

Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies