The eccentric angle of the point $\mathrm{P}(-6,2)$ of the ellipse $\frac{x^2}{48}+\frac{y^2}{16}=1$ is

The tangent to the ellipse $9 x^2+16 y^2=288$ making equal intercepts on the co-ordinate axes intersects the X -axis and the Y -axis in the points $A$ and $B$ respectively. Then $A(\triangle O A B)=$ (where O is origin)

The eccentricity of the curve represented by $x=3(\cos t+\sin t), y=4(\cos t-\sin t)$ is