The length of the chord of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$, whose mid-point is $\left(1, \frac{1}{2}\right)$, is :

Let the ellipse ${{\rm{E}}_1}:\frac{{{x^2}}}{{{{\rm{a}}^2}}} + \frac{{{y^2}}}{{\;{{\rm{b}}^2}}} = 1,{\rm{a}} > {\rm{b}}$ and ${{\rm{E}}_2}:\frac{{{x^2}}}{{\;{{\rm{A}}^2}}} + \frac{{{y^2}}}{{\;{{\rm{B}}^2}}} = 1,\;{\rm{A}} < {\rm{B}}$ have same eccentricity $\frac{1}{{\sqrt 3 }}$ ....

Let ${{\rm{E}}_1}:\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ be an ellipse. Ellipses ${{\rm{E}}_1}$'s are constructed such that their centres and eccentricities are same as that of ${{\rm{E}}_1}$,...

Let the product of the focal distances of the point $\left( {\sqrt 3 ,\frac{1}{2}} \right)$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,(a > b)$,...