$S=(-1,1)$ is the focus, $2 x-3 y+1=0$ 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 $3 a+2 b$ :

$a$ and $b$ are the semi-major and semi-minor axes of an ellipse whose axes are along the coordinate axes, If its latus rectum is of length 4 units and the distance between its foci is $4 \sqrt{2}$, then $a^{2}+b^{2}=$

If the extremities of the latus recta having positive ordinate of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a > b)$ lie on the parabola $x^{2}+2 a y-4=0$, then the points $(a, b)$ lie on the curve

The length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$, then $\sqrt{a^2+6 a b+b^2}=$