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

If the area of the auxiliary circle of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a > b} \right)$ is twice the area of the ellipse, then the eccentricity of the ellipse is

If $ \tan\  \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$  then the chord joining two points $\theta _1 \& \theta _2$  on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$  will subtend a right angle at :

A point on the ellipse, $4x^2 + 9y^2 = 36$, where the normal is parallel to the line, $4x -2y-5 = 0$ , is

If $PQ$ is a double ordinate of hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ such that $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies

An ellipse and a hyperbola have the same centre origin, the same foci and the minor-axis of the one is the same as the conjugate axis of the other. If $ e_1, e_2 $ be their eccentricities respectively, then  $e_1^{ - 2} + e_2^{ - 2}$ equals