The eccentricity of an ellipse is $2/3$,the length of the latus rectum is $5$,and the centre is $(0, 0)$. The equation of the ellipse is:

If $P$ lies in the first quadrant on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$),and the tangent and normal drawn at $P$ meet the major axis at points $T$ and $N$ respectively,then the value of $\frac{(\left| F_2N \right| + \left| F_1N \right|)(\left| F_2T \right| - \left| F_1T \right|)}{(\left| F_2N \right| - \left| F_1N \right|)(\left| F_2T \right| + \left| F_1T \right|)}$ is equal to (where $F_1$ and $F_2$ are the foci $(ae, 0)$ and $(-ae, 0)$ respectively).

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1$,where $a > 2$,having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis,be $6 \sqrt{3}$. Then the eccentricity of the ellipse is

The equation of the ellipse passing through the origin and having foci at $(1, 0)$ and $(3, 0)$ is .....

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{\pi}{6}$ sq. units. Which of the following is a possible equation of the ellipse?

If $\tan \theta_1 \times \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining $2$ points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at