Area (in sq. units) of the region outside $\frac{|\mathrm{x}|}{2}+\frac{|\mathrm{y}|}{3}=1$ and inside the ellipse $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is

An arch is in the form of a semi-cllipse. It is $8 \,m$ wide and $2 \,m$ high at the centre. Find the height of the arch at a point $1.5\, m$ from one end.

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to :

If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to

If the chord through the point whose eccentric angles are $\theta \,\& \,\phi $ on the ellipse,$(x^2/a^2) + (y^2/b^2) = 1$  passes through the focus, then the value of $ (1 + e)$ $\tan(\theta /2) \tan(\phi /2)$ is