Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 5,\,0),$ foci $(\pm 4,\,0)$

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

In a triangle $A B C$ with fixed base $B C$, the vertex $A$ moves such that $\cos B+\cos C=4 \sin ^2 \frac{A}{2} .$ If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$, respectively, then

$(A)$ $b+c=4 a$

$(B)$ $b+c=2 a$

$(C)$ locus of point $A$ is an ellipse

$(D)$ locus of point $A$ is a pair of straight lines

Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $  $ \&$  the diameter of the inscribed circle of triangle $OCF$  is $2, $ then the product $ (AB)\,(CD) $ is equal to

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 angle between the pair of tangents drawn to the ellipse $3{x^2} + 2{y^2} = 5$ from the point $(1, 2)$, is