The distance between the directrices of the ellipse $\frac{x^2}{36} + \frac{y^2}{20} = 1$ is

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(b < a)$ be an ellipse with major axis $AB$ and minor axis $CD$. Let $F_1$ and $F_2$ be its two foci,with $A, F_1, F_2, B$ in that order on the segment $AB$. Suppose $\angle F_1CB = 90^{\circ}$. The eccentricity of the ellipse is:

If the line $\alpha x+4y=\sqrt{7}$,where $\alpha \in R$,touches the ellipse $3x^{2}+4y^{2}=1$ at the point $P$ in the first quadrant,then one of the focal distances of $P$ is:

Find the equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$ and has an eccentricity of $\sqrt{2/5}$.

The sum of the focal distances of any point on the conic $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is

$A$ point $(4, -1)$ lies on an ellipse whose axes are parallel to the coordinate axes. If the line $x + 4y - 10 = 0$ is tangent to the ellipse at this point,find its equation $(a > b)$.