Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,
$$\phi \left( t \right) = {5 \over {12}} + t - {t^2}$$, then a² + b² is equal to :

The length of the minor axis (along y-axis) of an ellipse in the standard form is $${4 \over {\sqrt 3 }}$$. If this ellipse touches the line, x + 6y = 8; then its eccentricity is :

Let the line y = mx and the ellipse 2x² + y² = 1 intersect at a ponit P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$ and (0, $$\beta $$), then $$\beta $$ is equal to :

If 3x + 4y = 12$$\sqrt 2 $$ is a tangent to the ellipse
$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$$ for some $$a$$ $$ \in $$ R, then the distance between the foci of the ellipse is :