Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b < a)$, be a ellipse with major axis $A B$ and minor axis $C D$. Let $F_1$ and $F_2$ be its two foci, with $A, F_1, F_2, B$ in that order on the segment $A B$. Suppose $\angle F_1 C B=90^{\circ}$. The eccentricity of the ellipse is

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

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 5,\,0),$ foci $(\pm 4,\,0)$

The locus of the point of intersection of mutually perpendicular tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is

Let the foci and length of the latus rectum of an ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ be $( \pm 5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1$ equals