The equation $\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 5}} + 1 = 0$ represents an ellipse, if

If any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ cuts off intercepts of length $h$ and $k$ on the axes, then $\frac{{{a^2}}}{{{h^2}}} + \frac{{{b^2}}}{{{k^2}}} = $

An ellipse $\frac{\left(x-x_0\right)^2}{a^2}+\frac{\left(y-y_0\right)^2}{b^2}=1$, $a > b$, is tangent to both $x$ and $y$ axes and is placed in the first quadrant. Let $F_1$ and $F_2$ be two foci of the ellipse and $O$ be the origin with $OF _1 < OF _2$. Suppose the triangle $OF _1 F _2$ is an isosceles triangle with $\angle OF _1 F _2=120^{\circ}$. Then the eccentricity of the ellipse is

The equations of the common tangents to the ellipse, $ x^2 + 4y^2 = 8 $ $\&$  the parabola $y^2 = 4x$  can be

In the ellipse, minor axis is $8$ and eccentricity is $\frac{{\sqrt 5 }}{3}$. Then major axis is

The length of the latus rectum of an ellipse is $\frac{1}{3}$ of the major axis. Its eccentricity is