The circumcenter of the equilateral triangle having the three points $\theta_1, \theta_2, \theta_3$ lying on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its vertices is $(r, s)$. Then, the average of $\cos \left(\theta_1-\theta_2\right)$, $\cos \left(\theta_2-\theta_3\right)$ and $\cos \left(\theta_3-\theta_1\right)$ is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(b>a)$ is an ellipse with eccentricity $\frac{1}{\sqrt{2}}$. If the angle of intersection between the ellipse and parabola $y^2=4 a x$ is $\theta$, then the coordinates of the point $\frac{2 \theta}{3}$ on the ellipse is

If $P$ is any point on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $S, S^{\prime}$ are its foci, then the maximum area (in sq. units) of $\triangle S P S^{\prime}=$

Let $e$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

If $a=5, b=4$ and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is $l x+m y=27$ then $l+m=$