The eccentricity of the ellipse $\frac{(x - 1)^2}{9} + \frac{(y + 1)^2}{25} = 1$ is

Find the equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$ and has an eccentricity of $\sqrt{2/5}$.

The eccentricity of the ellipse $(x-3)^2 + (y-4)^2 = \frac{y^2}{9} + 16$ is -

If $(l, m)$ is the circumcentre of an equilateral triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ having vertices at points with eccentric angles $\theta_1, \theta_2$ and $\theta_3$,then $\frac{2}{3}\left[\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)+\cos \left(\theta_3-\theta_1\right)\right]=$

For the ellipse $25x^2 + 9y^2 - 150x - 90y + 225 = 0$,the eccentricity $e$ is: (in $/5$)

Let $P$ and $Q$ be the foci of an ellipse and let $R$ be one end of its minor axis. If $\triangle PQR$ is an equilateral triangle,then the eccentricity of the ellipse is equal to