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?

The equation of an ellipse, whose vertices are $(2, -2), (2, 4)$ and eccentricity $\frac{1}{3}$, is

In an ellipse $9{x^2} + 5{y^2} = 45$, the distance between the foci is

Let $S = 0$ is an ellipse whose vartices are the extremities of minor axis of the ellipse $E:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,a > b$ If $S = 0$ passes through the foci of $E$ , then its eccentricity is (considering the eccentricity of $E$ as $e$ )

A circle has the same centre as an ellipse and passes through the foci $F_1 \& F_2$  of the ellipse, such that the two curves intersect in $4$  points. Let $'P'$  be any one of their point of intersection. If the major axis of the ellipse is $17 $ and  the area of the triangle $PF_1F_2$ is $30$, then the distance between the foci is :

The eccentricity of the ellipse $25{x^2} + 16{y^2} - 150x - 175 = 0$ is