The eccentricity of the hyperbola whose length of the latus rectum is equal to $$8$$ and the length of its conjugate axis is equal to half of the distance between its foci, is :

Let $$P$$ be the point on the parabola, $${{y^2} = 8x}$$ which is at a minimum distance from the centre $$C$$ of the circle, $${x^2} + {\left( {y + 6} \right)^2} = 1$$. Then the equation of the circle, passing through $$C$$ and having its centre at $$P$$ is:

A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:

Consider :
f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$

A normal to $$y = $$ f$$\left( x \right)$$ at $$x = {\pi \over 6}$$ also passes through the point: