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

An ellipse having foci at $(3, 1)$ and $(1, 1)$ passes through the point $(1, 3)$. Then its eccentricity is

The eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

$S=(-1, 1)$ is the focus,$2x-3y+1=0$ is the directrix corresponding to $S$,and $\frac{1}{2}$ is the eccentricity of an ellipse. If $(a, b)$ is the centre of the ellipse,then $3a+2b=$

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the conic $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

In an ellipse,the distance between its foci is $6$ and the length of the minor axis is $8$. Then its eccentricity is