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 $\sqrt{a^2+6 a b+b^2}=$

$S$ is the focus of the ellips $\frac{x^2}{25}+\frac{y^2}{b^2}=1,(b<5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipes. If the distance between the foci of this ellipse is 8 and $S^{\prime} P=7$, then $\theta=$

The equations of the directrices of the elmpse $9 x^2+4 y^2-18 x-16 y-11=0$ are

$L_1^{\prime}$ is the end of a latus rectum of the ellipse $3x=2 \pm \frac{\sqrt{5}}{\sqrt{5}}$ $3 x^2+4 y^2=12$ which is lying in the third quadrant. If the normal drawn at $L_1^{\prime}$ to this ellipse intersects the ellipse again at the point $P(a, b)$, then $a=$