The coordinates of the foci of the ellipse $16x^{2} + 9y^{2} = 144$ are

In an ellipse,the distance between its foci is $6$ and the minor axis is $8$. Then its eccentricity is:

If $F_1$ and $F_2$ are the foci of the ellipse $16 x^2+25 y^2=400$ and $P$ is any point on it,then the value of the product $P F_1 \cdot P F_2$ lies in the interval

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:

If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its midpoint is $x+\alpha y=\beta$,then

$A$ tangent is drawn to the ellipse $\frac{x^2}{27} + y^2 = 1$ at the point $(3\sqrt{3} \cos \theta, \sin \theta)$ (where $\theta \in (0, \frac{\pi}{2})$). Then,the value of $\theta$ such that the sum of the intercepts on the axes made by this tangent is minimum,is