The extremities of the latera recta of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ having a given major axis $2a$ lie on:

The longest distance of the point $(a, 0)$ from the curve $2x^2+y^2=2x$ is

If the distance between the foci of an ellipse is equal to the length of the latus rectum, then its eccentricity 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 an ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ with vertices $A$ and $A'$,a tangent drawn at the point $P$ in the first quadrant meets the $y$-axis at $Q$,and the chord $A'P$ meets the $y$-axis at $M$. If $O$ is the origin,then $OQ^2 - MQ^2$ is equal to:

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is: