The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$ is

The area of the region enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is . . . . . . square units. (in $\pi$)

The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the coordinate axes is

The line $x = 8$ is the directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the corresponding focus $(2, 0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0, 4\sqrt{3})$ and intersects the $x$-axis at $Q$,then $(3PQ)^2$ is equal to $........$

The equations of the tangent and normal at point $(3, -2)$ of the ellipse $4x^2 + 9y^2 = 36$ are:

The equation of the normal drawn at the point $(\sqrt{2}+1, -1)$ to the ellipse $x^2+2y^2-2x+8y+5=0$ is