The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at

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

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

The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes.

Another ellipse $E _2$ passing through the point $(0,4)$ circumscribes the rectangle $R$.. The eccentricity of the ellipse $E _2$ 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

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 $\frac{x^{2}}{4}+\frac{y^2} {25}=1$.