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

Let $A,B$ and $C$ are three points on ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$where line joing $A \,\,\&\,\, C$ is parallel to the $x-$axis and $B$ is end point of minor axis whose ordinate is positive then maximum area of $\Delta ABC,$ is-

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 normal at any point $P$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ meets the co-ordinate axes in $G$ and $g$ respectively, then $PG:Pg = $

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $4 x ^{2}+9 y ^{2}=36$

The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3 r )^{2}$ is equal to