On the ellipse $${{{x^2}} \over 8} + {{{y^2}} \over 4} = 1$$ let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 $$-$$ e²). A is :

A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity $${1 \over 3}$$ and the distance of the nearer focus from this directrix is $${8 \over {\sqrt {53} }}$$, then the equation of the other directrix can be :

If a tangent to the ellipse x² + 4y² = 4 meets the tangents at the extremities of it major axis at B and C, then the circle with BC as diameter passes through the point :

Let an ellipse $$E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $${a^2} > {b^2}$$, passes through $$\left( {\sqrt {{3 \over 2}} ,1} \right)$$ and has eccentricity $${1 \over {\sqrt 3 }}$$. If a circle, centered at focus F($$\alpha$$, 0), $$\alpha$$ > 0, of E and radius $${2 \over {\sqrt 3 }}$$, intersects E at two points P and Q, then PQ² is equal to :