Which of the following points lies on the locus of the foot of perpedicular drawn upon any tangent to the ellipse,
$${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$
from any of its foci?

If the co-ordinates of two points A and B
are $$\left( {\sqrt 7 ,0} \right)$$ and $$\left( { - \sqrt 7 ,0} \right)$$ respectively and
P is any point on the conic, 9x² + 16y² = 144, then PA + PB is equal to :

Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is $${1 \over 2}$$. If P(1, $$\beta $$), $$\beta $$ > 0 is a point on this ellipse, then the equation of the normal to it at P is :

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 :