The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :

Let $\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1$ represent an ellipse with major axis along the y-axis,where $f$ is a strictly decreasing positive function on $R$. If the set of all possible values of $a$ is $R - [\alpha, \beta]$,then $\alpha^2 + \beta^2$ is equal to:

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

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$.

The center of the ellipse $x^2+2y^2-4x+12y+14=0$ is

If $ax^2 + by^2 = 15$ is the equation of the ellipse for which the distance between its foci is $2$ and the distance between its directrices is $5$,then $a + b =$