The centre of the ellipse $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$ is

The equation of the tangent to the ellipse $x^2+16y^2=16$ which makes an angle $60^{\circ}$ with the $X$-axis is

Let a focus of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $S(4, 0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3, \alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is equal to: (in $/ 5$)

For $0 < \theta < \frac{\pi}{2}$,four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents,the minimum value of $A(\theta)$ is

If $a$ and $c$ are positive real numbers and the ellipse $\frac{x^2}{4c^2} + \frac{y^2}{c^2} = 1$ has four distinct points in common with the circle $x^2 + y^2 = 9a^2$,then

If a circle $(x-1)^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally,then $r=$