Consider the parabola $${y^2} = 8x$$. Let $${\Delta _1}$$ be the area of the triangle formed by the end points of its latus rectum and the point $$P\left( {{1 \over 2},2} \right)$$ on the parabola and $${\Delta _2}$$ be the area of the triangle formed by drawing tangents at $$P$$ and at the end points of the latus rectum. Then $${{{\Delta _1}} \over {{\Delta _2}}}$$ is

Let $$f\left( \theta \right) = \sin \left( {{{\tan }^{ - 1}}\left( {{{\sin \theta } \over {\sqrt {\cos 2\theta } }}} \right)} \right),$$ where $$ - {\pi \over 4} < \theta < {\pi \over 4}.$$

Then the value of $${d \over {d\left( {\tan \theta } \right)}}\left( {f\left( \theta \right)} \right)$$ is

Let the eccentricity of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be reciprocal to that of the ellipse $${x^2} + 4{y^2} = 4$$. If the hyperbola passes through a focus of the ellipse, then

Let the straight line $$x=b$$ divide the area enclosed by
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals