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

For the ellipse $25x^2 + 9y^2 - 150x - 90y + 225 = 0$,the eccentricity $e$ is: (in $/5$)

If the tangent to the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is parallel to the $x$-axis,then what is the point of tangency?

If the eccentricity of the ellipse $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$ is $\frac{1}{\sqrt{6}}$,find the length of the latus rectum of the ellipse.

$A$ point moves such that the sum of its distances from $(ae, 0)$ and $(-ae, 0)$ is $2a$. Then the equation to its locus,where $b^2 = a^2(1 - e^2)$,is

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