Let the equations of two ellipses be ${E_1}:\,\frac{{{x^2}}}{3} + \frac{{{y^2}}}{2} = 1$ and ${E_2}:\,\frac{{{x^2}}}{16} + \frac{{{y^2}}}{b^2} = 1,$ If the product of their eccentricities is $\frac {1}{2},$ then the length of the minor axis of ellipse $E_2$ is

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ from any of its foci?

Find the equation for the ellipse that satisfies the given conditions: Centre at $(0,\,0),$ major axis on the $y-$ axis and passes through the points $(3,\,2)$ and $(1,\,6)$

An ellipse $\frac{\left(x-x_0\right)^2}{a^2}+\frac{\left(y-y_0\right)^2}{b^2}=1$, $a > b$, is tangent to both $x$ and $y$ axes and is placed in the first quadrant. Let $F_1$ and $F_2$ be two foci of the ellipse and $O$ be the origin with $OF _1 < OF _2$. Suppose the triangle $OF _1 F _2$ is an isosceles triangle with $\angle OF _1 F _2=120^{\circ}$. Then the eccentricity of the ellipse is

The distance between the foci of the ellipse $3{x^2} + 4{y^2} = 48$ is