If the focus of an ellipse is $(-1, -1)$,the equation of its directrix corresponding to this focus is $x + y + 1 = 0$,and its eccentricity is $e = \frac{1}{\sqrt{2}}$,then the length of its major axis is

The equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$,and has an eccentricity $e = \sqrt{\frac{2}{5}}$ is:

If $P$ lies in the first quadrant on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$),and the tangent and normal drawn at $P$ meet the major axis at points $T$ and $N$ respectively,then the value of $\frac{(\left| F_2N \right| + \left| F_1N \right|)(\left| F_2T \right| - \left| F_1T \right|)}{(\left| F_2N \right| - \left| F_1N \right|)(\left| F_2T \right| + \left| F_1T \right|)}$ is equal to (where $F_1$ and $F_2$ are the foci $(ae, 0)$ and $(-ae, 0)$ respectively).

What is the equation of the ellipse with foci $(\pm 2, 0)$ and eccentricity $e = \frac{1}{2}$?

The eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

If the distance between the foci of an ellipse is $6$ and the distance between its directrices is $12$,then the length of its latus rectum is